Assume not. It is equivalent to \[ a^2 + \frac{4a^2+e}{b} = c^2 \]for some $0 \le e < b$. Write this as \[ (b+4)a^2 - bc^2 = -e. \]At face, this is a cool assertion that a large family of Pell-like equations has no solutions. However it turns out we can Vieta jump by letting $c = \frac{x+y}{2}$ and $a = \frac{x-y}{2}$ (since $c > a$), which rewrites the equation as \[ x^2 - (b+2)xy + y^2 = -e. \]Then given $x \ge y \ge 1$ we have a Vieta jump \[ (x,y) \longmapsto \left( y, \frac{y^2+e}{x} \right) = \left( y, (b+2)y-x \right). \]To complete the descent argument, we need to check: Claim: For $xy \ge 2$, we always have $0 < \frac{y^2+e}{x} < x$. Proof. The left inequality is obvious. For right inequality, the given equation is \[ (x-y)^2 - bxy = -e > -b \implies (x-y)^2 > bxy-b \ge b \]and thus \[ x^2-y^2 = (x-y)(x+y) > (x-y)^2 > b > e \]which implies the desired result. $\blacksquare$ Therefore, the Vieta jumping can only conclude when $(x,y) = (1,1)$. However, plugging this back into the initial equation gives $2-(b+2) = -e$ or $b=e$, which doesn't work.

A very long solution using continued fractions: We try to bound \[ \left\lfloor a \sqrt{1 + 4/b} \right\rfloor^2 < a^2 + \left\lceil \frac{4a^2}{b} \right\rceil < \left\lceil a \sqrt{1 + 4/b} \right\rceil^2 . \]Since $a \sqrt{1+4/b}$ is obviously not an integer, this solves the problem. The lower bound is immediate from \[ \left\lfloor{a \sqrt{1+4/b}}\right\rfloor^2 < (a \sqrt{1 + 4/b})^2 \leq a^2 + \left\lceil{\frac{4a^2}{b}}\right\rceil . \]The upper bound is the difficult part. Let $\delta = \left\lceil{a \sqrt{1 + 4/b}}\right\rceil - a\sqrt{1 + 4/b}$. Then \[ \left\lceil{a \sqrt{1+ 4/b}}\right\rceil^2 = a^2 + \frac{4a^2}{b} + 2\delta a \sqrt{1 + 4/b} + \delta^2. \tag{$\dagger$} \]If $b$ divides $4a^2$, then the bound is immediate since $\left\lceil{\frac{4a^2}{b}} \right\rceil = \frac{4a^2}{b}$. Otherwise, it suffices to prove that \[ \delta \geq \frac{1}{2a} \iff \{ a\sqrt{1 + 4/b} \} \leq 1 - \frac{1}{2a} . \]This is where we devote most of our efforts to. Now we show that there are no solutions when $b \geq 5$. First, observe that if $b \in \{1,2,4 \}$, then $b \mid 4a^2$, so there are no solutions. Moreover if $b = 3$ and $3 \mid a$ there are no solutions, too. Otherwise $4a^2 \equiv 1 \pmod{3}$. Thus the expression is equal to a perfect square $c^2$ if and only if \[ 3c^2 - 7a^2 = 2. \]However, by taking both sides (mod 4), this is impossible. Henceforth, assume that $b \geq 5$. We shall use extensively various facts about continued fractions and diophantine approximations. Claim. Let $\lambda = \frac{1}{2}({\sqrt{1 + 4/b}} - 1)$. Then \[ \lambda = [0, b, 1, b, 1, \dots ], \]using the standard notation for simple continued fractions. Proof. It is well known that \[ \sqrt{x} = 1 + \dfrac{x-1}{2 + \frac{x-1}{2 + \frac{x-1}{2+ \cdots}}}. \]Inputting $x = {1 + b/4}$, we obtain \begin{align*} \sqrt{1 + 4/b} &= 1 + \frac{4/b}{2 + \frac{4/b}{2 + \frac{4/b}{2 + \frac{4/b}{ 2 + \cdots}}}} \\ &= 1 + \frac{4}{2b + \frac{4}{2 + \frac{4}{2b + \frac{4}{2+ \cdots}}}} \\ &= 1 + \frac{2}{b + \frac{1}{1 + \frac{1}{b + \frac{1}{1 + \cdots}}}}. \end{align*}The claim follows immediately. Then \[ \{ a \sqrt{1 + 4/b} \} = \{ 2a \lambda \}. \]This is what we will bound. Consider the sequence of convergents $(p_n/q_n)$ that come from the continued fraction of $\lambda$. We use three well known facts about the convergents for a simple continued fraction $[a_0, a_1, a_2, \dots ]$. If $n$ is odd, then $p_n/q_n$ is greater than $\lambda$. If $n$ is even, then $p_n/ q_n$ is less than $\lambda$. For each $n \geq 2$, \[ q_n = a_n q_{n-1} + q_{n-2} = \begin{cases} q_{n-1} + q_{n-2} &\text{if $n$ is even,} \\ bq_{n-1} + q_{n-2} &\text{if $n$ is odd.} \end{cases} \tag{1} \]As a corollary of this, we have that the sequence $(q_n)$ is strictly increasing. Moreover, because $q_1 = b$, we also have that $b$ divides $q_n$ whenever $n$ is odd by induction. For each $n$, \[ \left| { \lambda - \frac{p_n}{q_n}} \right| < \frac{1}{q_n q_{n+1}} . \tag{2} \] For each $n$, \[ p_n q_{n-1} - p_{n-1} q_n = (-1)^{n-1}. \tag{3} \] Let $n$ be the greatest index such that $q_n \leq 2a < q_{n+1}$. Let $\varepsilon = | \lambda - p_n / q_n |$ and $\hat{a} = 2ap_n \pmod{q_n}$. It follows that \[ \{ 2a \lambda \} = \left\{ \frac{\hat{a}}{q_n} \pm 2a \varepsilon \right\}, \]with the sign $\pm$ depending on the parity of $n$. By (2), we have that \[ 2a \varepsilon < \frac{2a}{q_n q_{n+1}} < \frac{1}{q_n}. \]This gives us almost everything we need. From here, we split into cases based on the parity of $n$. Case 1.If $n$ is odd, then as long as $\hat{a} \neq 0$, then \[ 0 < \{ 2a \lambda \} < \frac{q_n-1}{q_n} < 1 - \frac{1}{2a}. \]Thus the only problem is if $q_n \mid 2a$ (bearing in mind that $\gcd(p_n, q_n) = 1$). However, by assumption $b$ does not divide $2a$, so this is impossible since $b$ divides $a_n$. Case 2. If $n$ is even, then we have \[ \{2a \lambda \} < \frac{q_n-2}{q_n} + \frac{1}{q_n} < 1 - \frac{1}{2a} \]unless $\hat{a} = q_n - 1$. Thus we just have to deal with this edge case. First, we will deal with if $2a > q_{n+1} - 1.5q_n$.Then we will prove that \[ \frac{1}{x} + \frac{x}{q_n q_{n+1}} < \frac{1}{q_n} \tag{4} \]for all $x \in [q_n, q_{n+1} - 1.5q_n]$. Taking $x = 2a$, this would imply that \[ \{ 2a \lambda \} < 1 - \frac{1}{q_n} + \frac{2a}{q_nq_{n+1}} < 1 - \frac{1}{2a}. \] Suppose that $2a > q_{n+1} - 1.5q_n$. Because $2a$ is very close to $q_{n+1}$, we may use $\frac{p_{n+1}}{q_{n+1}}$ as a very good diophantine approximation. More precisely, we have \[ \{ 2a \lambda\} < 1 - \frac{1}{q_{n+1}} - \frac{2a}{q_{n+`1} q_{n+2}} \]Thus it is sufficient to prove that \[ \frac{1}{2a} - \frac{2a}{q_{n+1} q_{n+2}} < \frac{1}{q_{n+1}} \]From here, we invoke a number of bounds. By applying (1), we obtain that $q_{n} < \frac{1}{b} q_{n+1}$. Thus $2a \geq q_n( 1 - \frac{3}{2b})$ Again from (1), we have that $q_{n+2} < 2q_{n+1}$. Thus \[ \frac{1}{2a} - \frac{2a}{q_{n+1} q_{n+2}} < \frac{1}{q_n(1 - \frac{3}{2b})} - \frac{q_{n+1}}{2q_{n+1}^2} = \frac{1}{q_{n+1}} \left( \frac{2b}{2b-3} - \frac{1}{2} \right) \]For $b \geq 5$, we have $\frac{2b}{2b-3} - \frac{1}{2} < 1$, so we are done. Now we just have to do some analysis to prove the bound in (4). Let $f(x) = \frac{1}{x} + \frac{x}{q_nq_{n+1}}$. Then $f'(x) = \frac{1}{q_nq_{n+1}} - \frac{1}{x^2}$. Hence $f'$ changes sign just once on $[q_n, q_{n+1} - 1.5 q_n]$. Our strategy is to find prove that $f(q_{n+1} - 1.5 q_n) < \frac{1}{q_n}$ and to find $\alpha \leq 2a$ such that $f( \alpha) < \frac{1}{q_n}$. Because $f'$ changes sign just once, this is enough to imply the inequality in $(4)$. The bound for $f(q_{n+1} - 1.5 q_n)$ is just a computaiton: \[ f(q_{n+1} - 1.5 q_n) = \frac1{q_{n+1} - 1.5 q_n} + \frac{1}{q_n} - \frac{3}{2q_{n+1}} . \]Using that $q_n < \frac{1}{b} q_{n+1}$ we may bound \[ \frac{1}{q_{n+1} - 1.5 q_n} < \frac{1}{q_{n+1}(1 - \frac{3}{2b})} < \frac{3}{2} q_{n+1} \]with the final inequality holding for $b \geq 5$. This establishes $f(q_{n+1} - 1.5 q_n) < \frac{1}{q_n}$. I claim we can take $\alpha = \frac{3}{2} q_n$. By (3), \[ 2a \equiv q_{n+1} \pmod{q_n} \]But \[ q_{n+1} \pmod{q_n} = q_{n-1} > \frac{q_n}{2} \]by (1). Thus $2a \geq \frac{3}{2} q_n$. Again, we check that \[ f \left( \frac{3}{2} q_n \right) = \frac{2}{3q_n} + \frac{3}{2q_{n+1}} < \left( \frac{2}{3} + \frac{3}{2b} \right) \frac{1}{q_n} < \frac{1}{q_n}. \]This concludes the proof.

PEN A4 nnosipov wrote: If $a$, $b$, $c$ are positive integers such that \[ |a^2+b^2-abc-2|<c, \]then $a^2+b^2-abc$ is a perfect square.

v_Enhance wrote: At face, this is a cool assertion that a large family of Pell-like equations has no solutions. (Proposed by me ).Yes, one could use this old classical idea which was hidden in the new frames. But I think there are many different approaches.

The same substitution also appeared in Shortlist 2016 N5.

Comment: This is a neat problem. However, the trick of multiplying units to descent is somewhat known. Had I thought about this during the TST, I would have solved it easily. Replace $b$ with $k$. First, we reduce the equation to a general Pell's equation. Claim: The equation implies $k(k+4)x^2-y^2 \in\{0,4,8,\hdots,4k-4\}$ for some $x,y\in\mathbb{Z}^+$ Proof. Set $4a^2\equiv -r\pmod k$ where $r\in\{0,1,\hdots,k-1\}$. Then we have (for some $b\in\mathbb{Z}$) that $$ka^2 + (4a^2+r) = kb^2 \implies kb^2 - (k+4)a^2 = r.$$Motivated by the identity $$(\sqrt{k+4}-\sqrt{k})(b\sqrt{k}-a\sqrt{k+4}) = -(a(k+4)+bk)+(a+b)\sqrt{k(k+4)},$$we set $x = a+b$ and $y=a(k+4)+bk$; a straightforward algebra yields the claim. $\blacksquare$ Now we do the descent. Consider the solution with $x$ as small as possible. Multiplying by $\sqrt{k(k+4)}-k$, we get the miraculous identity $$ k(k+4)\left(y-(k+2)x\right)^2 - \left(k(k+4)x - (k+2)y\right)^2 = 4\left(k(k+4)x^2 - y^2\right).$$Thus, we have the solution (a straightforward parity check gives that these are indeed integers) $$\left(\frac{y-(k+2)x}{2}, \frac{k(k+4)x - (k+2)y}{2}\right).$$By minimality, we have $$\frac{y-(k+2)x}{2} \notin (-x,x) \iff y \notin (kx, (k+4)x)$$so we split into two cases. If $y\leq kx$, then $k(k+4)x^2 - y^2 \geq 4kx^2\geq 4k$, contradiction. If $y\geq (k+4)x$, then $k(k+4)x^2 - y^2 \leq k(k+4)x^2 - (k+4)^2x^2 < 0$, contradiction. Hence, we are done by infinite descent.

redacted

Recall that ELMO 2012 SL N6 states that for any positive integers $x$, $y$ with $(x, y) \ne (1, 1)$, we have \[\left\lfloor\frac{(x-y)^2-1}{xy}\right\rfloor=\left\lfloor\frac{(x-y)^2-1}{xy-1}\right\rfloor.\]Now return to the original problem. Suppose that the expression equals $(a+k)^2$, with $k > 0$. Apply the ELMO SL problem to $(k, 2a+k)$ to obtain that \begin{align*} \left \lfloor \frac{4a^2 - 1}{2ak + k^2} \right \rfloor & = \left \lfloor \frac{4a^2 - 1}{2ak + k^2 - 1} \right \rfloor\\ \iff \left \lceil \frac{4a^2}{2ak + k^2} \right \rceil & = \left \lceil \frac{4a^2}{2ak + k^2 - 1} \right \rceil. \end{align*}However, $a^2 + \left \lceil \tfrac{4a^2}{b} \right \rceil = (a + k)^2$ implies \[\frac{4a^2}{2ak + k^2} \le b < \frac{4a^2}{2ak + k^2 - 1},\]contradiction.

CantonMathGuy wrote: Recall that ELMO 2012 SL N6 states that for any positive integers $x$, $y$ with $(x, y) \ne (1, 1)$, we have \[\left\lfloor\frac{(x-y)^2-1}{xy}\right\rfloor=\left\lfloor\frac{(x-y)^2-1}{xy-1}\right\rfloor.\]Now return to the original problem. Suppose that the expression equals $(a+k)^2$, with $k > 0$. Apply the ELMO SL problem to $(k, 2a+k)$ to obtain that \begin{align*} \left \lfloor \frac{4a^2 - 1}{2ak + k^2} \right \rfloor & = \left \lfloor \frac{4a^2 - 1}{2ak + k^2 - 1} \right \rfloor\\ \iff \left \lceil \frac{4a^2}{2ak + k^2} \right \rceil & = \left \lceil \frac{4a^2}{2ak + k^2 - 1} \right \rceil. \end{align*}However, $a^2 + \left \lceil \tfrac{4a^2}{b} \right \rceil = (a + k)^2$ implies \[\frac{4a^2}{2ak + k^2} \le b < \frac{4a^2}{2ak + k^2 - 1},\]contradiction. Wow, what a trick

indeed nice!

This was Greek TST Problem 4

I prove the stronger statement that $a^2+4\left\lceil\frac{a^2}{b}\right\rceil$ is not square. (This is stronger because replacing $a$ with $2a$ gives the original problem times 4.) I will use induction on $a$. If $a=1$, the sum is always 5, so we are done. Now assume it is true for $1,2,\cdots a-1$, and assume $a^2+4\left\lceil\frac{a^2}{b}\right\rceil=(a+d)^2$ for some $d$. Notice that $d$ must be even, so let $d=2k$. Then we have $\left\lceil\frac{a^2}{b}\right\rceil=ak+k^2$. Now notice that $$a^2-k^2<a^2\implies \frac{a-k}{k}<\frac{a^2}{k(a+k)}$$and $$ak\leq ak+k^2-1 \implies \frac{a^2}{ak+k^2-1}\leq \frac{a}{k}$$Thus, since $ak+k^2-1<\frac{a^2}{b}\leq ak+k^2$, we have $$\frac{a}{k}-1<\frac{a^2}{ak+k^2}\leq b<\frac{a^2}{ak+k^2-1}\leq \frac{a}{k}$$Let $a=nk+r$ for some $0\leq r\leq k-1$. If $n=0$, then $\left\lceil\frac{a^2}{b}\right\rceil\leq r^2<rk+k^2$, which is a contradiction. If $r=0$, then both $\frac{a}{k}-1$ and $\frac{a}{k}$ are integers, contradicting the above inequality. But then $n-1+\frac{r}{k}<b<n+\frac{r}{k}$, so $b=n$. Thus, we have $$\left\lceil\frac{a^2}{b}\right\rceil=\left\lceil\frac{n^2k^2+2nkr+r^2}{n}\right\rceil=nk^2+2kr+\left\lceil\frac{r^2}{n}\right\rceil$$And since $ak+k^2=k^2+nk^2+kr$, we get $k^2=kr+\left\lceil\frac{r^2}{n}\right\rceil$. Thus, $r^2+4\left\lceil\frac{r^2}{n}\right\rceil=r^2+4(k^2-kr)=(2k-r)^2$. But from induction, since $r<k<a$, $r^2+4\left\lceil\frac{r^2}{n}\right\rceil$ is not a square, so this is a contradiction. Thus, $a^2+4\left\lceil\frac{a^2}{b}\right\rceil$ is never a square.

Replace b with k well then we reduce it to a general Pells equation Claim: The equation implies $k(k+4)x^2-y^2 \in\{0,4,8,\hdots,4k-4\}$ for some x,y is an element of Z^+ Proof: Set 4a^2 is congruent -r(mod k) where r is an element of {0,1....,k-1}. Then we have (for some b is an element of Z) that ka^2+(4a^2+r)=kb^2, kb^2-(k+4)a^2=r Motivated by the identity (sqrtk+4-sqrtk)(bsqrtk-asqrtk+4)=-(a(k+4)+bk+(a+b)sqrtk(k+4), we can now set to x=a+b and y=a(k+4)+bk which is straightforward algebra which yields the claim. $\blacksquare$ Consider the solution with x as small as possible multiplying by sqrtk(k+4)-k we get the miraculous identity k(k+4) (y-(k+2)x^2-(k(k+4)x-(k+2)y^2=4(k(k+4)x^2-y^2). Thus we have a solution that is (a straightforward parity check gives that these are indeed integers) y-k(+2)x/2 isn't an element (-x,x) which is y is not an element of (kx,(k+4)x) so we split into two cases If y<=kx then k(k+4)x^2-y^2>=4kx^2>=4k, contradiction. If y>=(k+4)x, then k(k+4)x^2-y^2<=k(k+4)x^2-(k+4)x^2x^2<0 which is a contradiction. Hence it is a contradiction Note : I would personally do Evan Chens solution well related to his it is the vieta jumping one which is pretty clean ig this was a good ISL I guess :starwars. Credits to Markbcc186 or acknowledgements to him. I didn't write this up.

Posting for storage. Assume there exists a positive integer $c$ such that \[a^2+\left\lceil\frac{4a^2}b\right\rceil=c^2\]Hence \[\left\lceil\frac{4a^2}b\right\rceil=c^2-a^2=(c+a)(c-a)\]Let $c+a=x$ and $c-a=y$. Hence \[\frac{(x-y)^2+r}b=xy\]with $0\le r<b$. Of course $x\ne y$ so assume $x>y$. Moreover assume $x+y$ is minimal. The equation may be written as \[x^2-(b+2)xy+y^2+r=0\]Observe that if $(x,y)$ is a solution to the equation then $\left((b+2)y-x,y\right)$ is also a solution. Hence it suffices to prove that \[(b+1)y<x<(b+2)y\]to obtain a contradiction. It is true since \[0>y^2-(b+2)xy=y\left[x-(b+2)y\right]\]and \[0<x^2-(b+2)xy+y^2+b=(x-y)^2+b(1-xy)\le (x-y)^2+b(y^2-xy)=(x-y)(x-y-by)=(x-y)\left[x-(b+1)y\right]\]

This can also be treated as an algebra problem. First treat the case $b \ge a$. The only possibility is $a^2 + \lceil 4a^2/b \rceil = (a+1)^2$. We see that for $b = 2a$ the LHS is $a^2 + 2a < (a+1)^2$ and for $b = 2a-1$ it is $a^2 + 2a + 2 > (a+1)^2$, handling this case. We may hence assume $b < a$. Write $a^2 + \lceil 4a^2/b \rceil = (a+k)^2$ for $k \ge 0$ an integer, so in other words $k$ is an integer satisfying \begin{align*} 2ak + k^2 - 1 < 4a^2/b \le 2ak + k^2. \end{align*}Solving the two inequalities for $k$, after a bit of calculation we obtain that $k$ lies in the interval \begin{align*} (a(\sqrt{1 + 4/b + 1/a^2} - 1), a(\sqrt{1 + 4/b} - 1)]. \end{align*}We now use a Taylor expansion for $\sqrt{1 + x}$ around $x = 0$ to get $\sqrt{1+x} = 1 + x/2 + O(x^2)$. We remark that when near $0$, the sign of the $O$-term is negative. Now $k$ lies in \begin{align*} (2a/b + 1/a + O(1/b^2), 2a/b + O(1/b^2)] \ \ \ \ \ (\ast). \end{align*}This implies \begin{align*} |k - 2a/b| \le 1/a + O(1/b^2), \end{align*}so after multiplying by $b$ we get \begin{align*} |bk - 2a| \le b/a + O(1/b). \end{align*}The idea is that the right hand side should be strictly less than $1$, after which one concludes that $b \mid 2a$. This requires a bit of more detail in the calculations (checking, for example, the case $b > a/2$, and bounding the implied constant in the $O$-term), which we omit here. Now that one has $b \mid 2a$, one returns to $(\ast)$ and notes that since the sign of the $O$-term is negative, the relevant interval in fact does not contain $k$.

Sorry for no LaTeX (pic) and Russian language.

adilbek wrote: Sorry for no LaTeX (pic) and Russian language. Our team solution.

Alga Kyrgyzstan!!!

naman12 wrote: Let $a$ and $b$ be two positive integers. Prove that the integer \[a^2+\left\lceil\frac{4a^2}b\right\rceil\]is not a square. (Here $\lceil z\rceil$ denotes the least integer greater than or equal to $z$.) Russia Solution. Clearly $b=1,2,4$ are not possible. It is not hard to handle the case $b=3.$ So we may assume that $b>4$ hereon. If $b=2b'$ is even then it is not hard to check that we must have $2a^2>b'$ and if $a=a_0$ works then $a=(n-a_0)b'-a_0$ also works. Suppose $a_0$ is the minimum working $a$. Then we must have $a_0\le (n-a_0)b'-a_0$, which after some simplification, reduces to $2a_0n\le r<b.$ Absurd. From now on assume $b$ is odd. Set $b+2=m.$ We are basically trying to show that $bn^2-(b+4)a^2=r$ has no solutions in positive integers if $0\le r<b.$ Assume the contrary that $(n,a)$ is a solution of $bn^2-(b+4)a^2=r$ such that $n$ is minimal. Consider the Pell's equation $$x^2-b(b+4)y^2=1\iff x^2-(m^2-2)y^2=1,$$it is easy to verify that $(x_0,y_0)=\left(\tfrac12 m(m^2-3),\tfrac12 (m^2-1)\right)$ is a solution. It is easily verified that $(x,y)=(|nx_0-(m+2)ay_0|,|nby_0-ax_0|)$ is a solution of $bx^2-(b+4)y^2=r$. First of all, $nx_0\ge (m+2)ay_0$ because $$\left(\frac na\right)^2=1+\frac{4}{m-2}+\frac{r}{a^2(m-2)}\ge \frac{m+2}{m-2}\ge \frac{(m^2-1)(m+2)}{m(m^2-3)}=\frac{(m+2)y_0}{x_0}.$$By minimality of $n$ it follows that \begin{align*} &n\le nx_0-(m+2)ay_0=\frac12(m(m^2-3)n-(m+2)(m^2-1)a)\\ \implies & n\ge \frac{a(m+2)(m^2-1)}{m(m^2-3)-2}=\frac{a(m-1)(m+2)}{(m-2)(m+1)} \end{align*}Squaring and substituting $n^2=a^2\left(1+\frac{4}b+\frac{r}{a^2b}\right)$ we obtain, $$1+\frac{4}{m-2}+\frac{r}{a^2(m-2)}\ge \left(\frac{(m-1)(m+2)}{(m-2)(m+1)}\right)^2$$Solving for $a^2,$ $$a^2\le \frac{r(m-2)(m+1)^2}{4(m+2)}\implies a\le \frac{b(b+3)}{2\sqrt{b+4}}$$Obviously $n=\left\lceil a\sqrt{1+\frac 4b}\right\rceil.$ Set $x=\tfrac 4b.$ By Taylor Series, $$\sqrt{1+x}=1+\frac x2-\frac{x^2}{8}+\cdots$$Therefore, $$\left|\sqrt{1+x}-\left(1+\frac x2\right)\right|\le \frac18(x^2+x^3+\cdots)=\frac{x^2}{8(1-x)}$$Whence, $$\left|a\sqrt{1+\frac 4b}-\left(a+\frac{2a}{b}\right)\right|\le \frac{2a}{b(b-4)}\le\frac{b(b+3)}{2\sqrt{b+4}}\le \frac{8}{\sqrt{b+4}}<1.$$So it follows that $$n\in\left\{\left\lceil a+\frac{2a}{b}\right\rceil, \left\lceil a+\frac{2a}{b}\right\rceil-1\right\}$$because $\sqrt{1+x}<1+\tfrac x2.$ Let $\ell$ be the least non-negative integer such that $b\mid 2a+\ell,$ remark that $b\mid \ell^2+r.$ If $n=a+\tfrac{2a+\ell}{b}$ then $b$ comes out to be $-\frac{(2a+\ell)^2}{2a\ell-r}.$ This forces $r>2a\ell$ but, $r-2a\ell=\ell^2+r-\ell(2a+\ell)$ is divisible by $b$ hence $r-2a\ge b$ but on the other hand $r<b.$ Absurd. Now for the second case, let $\ell$ be the smallest non-negative integer such that $n=a+\frac{2a-\ell}{b}.$ Again we remark that $b\mid \ell^2+r.$ Putting this expression for $n$ and solving for $b$ we obtain $$b=\frac{(2a-\ell)^2}{2a\ell+r}.$$Let $2a=bq+\ell$ where $q$ is the quotient when $2a$ is divided by $b.$ Simplifying the above, $$q^2-q\ell-\left(\frac{\ell^2+r}{4}\right)=0$$Solving for $q$ we get $q=\frac12(b\ell\pm x)$ where $$\ell^2+4\left(\frac{\ell^2+r}{b}\right)=x^2.$$Now, $$\ell^2<x^2=\ell^2+\frac{4(\ell^2+r)}{b}<\ell^2+\frac{4\ell^2}{b}+\frac{4r}{b}<\ell^2+4\ell+4\implies x=\ell+1,$$plugging we get $4\mid 2\ell+1$ which is a contradiction. And we are done. $\blacksquare$

For contradiction, suppose we can write \[a^2+\lceil \tfrac{4a^2}{b}\rceil = c^2\]for some integer $c$. Let $-4a^2$ be $e\pmod{b}$ for $0\le e<b$, then the equation rewrites as \[a^2+\frac{4a^2+e}{b}=c^2\implies (b+4)a^2+e=bc^2\implies (b+4)a^2-bc^2=-e.\]Let $a=\frac{1}{2}(x-y),c=\frac{1}{2}(x+y)$ (with positive $y$ because $c>a$) for integers $x,y$, then \[x^2-(b+2)xy+y^2=-e.\]We now Vieta Jump. WLOG, $x\ge y\ge 1$, so we jump \[(x,y)\mapsto \left(y,\frac{y^2+e}{x}\right)=(y,(b+2)y-x).\]If $y>\frac{y^2+e}{x}$, our jump is successful, so suppose we jump until $y\le \frac{y^2+e}{x}$. Then, we have \[y\le x\le y+\frac{e}{y}.\]Define $f(x)=x^2-(b+2)xy+y^2$ and note $f'(x)=2x-(b+2)y$, so $f$ has absolute minimum at $x=(b+2)y/2$. Observe the bound \[(b+2)y/2 - y-e/y = by/2-e/y> by/2-b/y = b(y/2-1/y)\ge b(y/2-1).\]This bound implies $f'(x)<0$ when $y>1$. We now resolve the case $y=1$. We have $x^2-(b+2)x+1=-e$, so $x^2-(b+2)x+1+e=0$. Hence, we have \[x=\frac{b+2\pm \sqrt{b^2+4b+4-4-4e}}{2} = \frac{b+2\pm\sqrt{b^2+4b-4e}}{2},\]so because $x$ is an integer, $\sqrt{b^2+4b-4e}$ is an integer. Let $\sqrt{b^2+4b-4e}=r$, then $r^2-b^2=4(b-e)$, but $0<4(b-e)<4b+4$ implies $r=b+1$, contradiction. Hence, the fact that $x^2-(b+2)xy+y^2 \le y^2-(b+2)y^2+y^2=-by^2<-e$ finishes, so we may continue our Vieta Jumping until exactly one of $x,y$ is zero (since there must be some first zero $x,y$). But if $x=0$, $y^2=-e$ implies $y$ is zero and vice versa, so we have a contradiction and there was no initial pair $a,b$.

Solution (Similar to other solutions above): We will proceed by contradiction, so we have, for some $0 \leq d <b$, that $a^2 + \frac{4a^2+d}{b} = c^2$ for some positive integer $c$. Note that $c$ is greater than $a$. Rearranging yields $c^2b - (a^2b+4a^2) = d \implies 4a^2 + b(a-c)(a+c) = -d$. So now if we let $m = c+a$ and $n = c-a$ (We can easily see that $1 \leq n <m$), we get $bmn - 4(\frac{m+n}{2})^2 = d \implies m^2+n^2-(b-2)*mn = -d$. Now is the time we Vieta Jump (as others have done). So, $(m,n)\mapsto (m_0, n) = \left(\frac{n^2+d}{m}, n\right)$. From this, we know $m_0 = \frac{n^2+d}{m}$ is a positive integer since it can also be written as $b-(m+2)$ (which is obviously a positive integer). Thus, if we can get that $m_0 <m$, then the descent is confirmed. So if that were to be true, then $n^2+d <m^2$. However, notice that $m^2-n^2 = 4ac>4a^2$ (since $c>a$) $ = b^2(c^2-a^2) -d > b$ and so it would clearly be greater than $d$ since $c^2-a^2 >2$, as claimed. $\square$

We prove the stronger result (which I see was also proven in post #15 ) that there aren't positive integers $a,b,k$ such that $$\dfrac{a^2}{4}+\left\lceil\dfrac{a^2}{b}\right\rceil=\dfrac{k^2}{4}$$Call this equation $(1)$. The problem follows from the particular case in which $a$ and $k$ are even. We proceed by contradiction, that is, assume that there exist positive integers $a,b,k$ such that $(1)$ holds. Choose the solution for $(1)$ in which $a+k$ is minimum. As $\dfrac{a^2}{4}+\left\lceil\dfrac{a^2}{b}\right\rceil>\dfrac{a^2}{4}$, for the RHS of $(1)$ to be a perfect square divided by 4, we must have that $$\dfrac{a^2}{4}+\dfrac{a^2}{b}\ge \dfrac{a^2}{4}+\left\lceil\dfrac{a^2}{b}\right\rceil\ge \dfrac{(a+1)^2}{4}>\dfrac{a^2+2a}{4}$$that is, \begin{align*} \dfrac{a^2}{4}+\dfrac{a^2}{b}&> \dfrac{a^2}{4}+\dfrac{a}{2}\\ \dfrac{a^2}{b}&> \dfrac{a}{2}\\ 2a&> b\\ \end{align*}Now, by the Division Algorithm, we can find nonnegative integers $q,r$ such that $$a^2=qb-r$$and $0\le r< b$. Call this equation $(2)$. Plugging $(2)$ in $(1)$ and making the sustitution $x=\dfrac{k+a}{2}$ and $y=\dfrac{k-a}{2}$, we get \begin{align*} \dfrac{a^2}{4}+\left\lceil\dfrac{a^2}{b}\right\rceil &=\dfrac{k^2}{4}\\ \dfrac{a^2}{4}+\left\lceil\dfrac{qb-r }{b}\right\rceil &=\dfrac{k^2}{4}\\ \dfrac{a^2}{4}+q&=\dfrac{k^2}{4}\\ q&=\bigg(\dfrac{k+a}{2}\bigg)\bigg(\dfrac{k-a}{2}\bigg)\\ q&=xy\\ \end{align*}Here, note that integers $k+a$ y $k-a$ have the same parity. Now, as $q$ is an integer, both must be even. This proves that $x,y$ ar positive integers with $x>y$. ($k>a$ by $(1)$, as the ceil of a positive number is at least 1). Now, plugging this in $(2)$ and noting that $a=x-y$, we get \begin{align*} (x-y)^2&=xyb-r\\ x^2-2xy+y^2&=xyb-r\\ x^2-(2y+by)x+y^2+r&=0\\ \end{align*}Then, $x$ is a root of the equation $$X^2-(2y+by)X+y^2+r=0$$Call this equation $(3)$. Let $x'$ be the other (possibly doble, for now) root of $(3)$. Note that, as $k\ge 2$ (because the LHS in $(1)$ is at least 1), by Vieta, $$x'=\dfrac{y^2+r}{x}<\dfrac{y^2+b}{x}<\dfrac{y^2+2a}{x}\le\dfrac{y^2+ak}{x}=\dfrac{y^2+(x^2-y^2)}{x}=x$$As $x+x'=2y+by$, by Vieta, too, $x'$ is integer. In few words, $x'$ is an integer such that $0<x'<x$ and also satisfies $(3)$. To conclude, if we say $a'=|x'-y|$ (which is positive, because plugging $X=y$ in $(3)$ yields $r=y^2b>r$, a contradiction), $q'=x'y$ and $k'=x'+y$, as we still have that $0\le r<b$ and $x',y$ satisfy $(3)$, which can be rewritten as $(X-y)^2=Xyb-r$, we have \begin{align*} q'&=x'y\\ q'&=\dfrac{(x'+y+|x'-y|)(x'+y-|x'-y|)}{4}\\ q'&=\dfrac{(k'+a')(k'-a')}{4}\\ \dfrac{a'^2}{4}+q'&=\dfrac{k'^2}{4}\\ \dfrac{a'^2}{4}+\left\lceil\dfrac{q'b-r }{b}\right\rceil &=\dfrac{k'^2}{4}\\ \dfrac{a'^2}{4}+\left\lceil\dfrac{a'^2}{b}\right\rceil &=\dfrac{k'^2}{4}\\ \end{align*}Then, $a',b,k'$ are positive integers satisfying $(1)$, but $$a'+k'=|x'-y|+x'+y= 2\max\{x',y\}<2x=a+k$$and we chose $a+k$ minimum. Contradiction! Hence, we are done. $\square$ Remark. I found the generalization because I needed it for my Vieta Jumping to work.

Assume that the expression gives a square for some $a$ and $b$. Thus we can represent the expression as: $$a^2 + \left\lceil \frac{4a^2}{b} \right\rceil = k^2$$Obviously we have that $k=a+t$, thus giving us that: $$\left\lceil \frac{4a^2}{b} \right\rceil =2at+t^2$$Since we know for certainty that $ \frac{4a^2}{b} \leq \left\lceil \frac{4a^2}{b} \right\rceil \leq \frac{4a^2}{b}+1$ When we plug in the inequality we must have that: $$b \leq \frac{4a^2}{2at+t^2-1} \leq 2a$$ Now let's say that $\left\lceil \frac{4a^2}{b} \right\rceil = \frac{4a^2+p}{b}$, where $p$ is a positive integer and less or equal to than $b$. When plugged in we get that: $$(b+4)a^2-bk^2=-p$$ Since $k > a$, we can represent them as $k=\frac{x+y}{2}$ and $a=\frac{x-y}{2}$. Plugging in now whate we got we have that: $$x^2-(b+2)xy+y^2=-p$$This screams of Vieta jumping,so let's try it. Let $S$ be the set of pairs of integers $(x,y)$ such that the above expression holds. Now let $(X,Y)$ be the minimal pair, i.e. their sum is minimal. Now we shall search for a substitutions, i.e. when does the equation still hold under some transformation. We easily find by reverse constructing that the pair $(y,(b+2)y-x)$ satisfies the equation. But notice that $(b+2)y-x= \frac{y^2+p}{x}$, thus we wnat to find the piars for which we have that: $$\frac{y^2+p}{x}<x$$i.e. when the equation fails to holds under Vieta jumping. The following inequality is equivalent to the one above: $$(b+2)xy < 2x^2$$Notice that we have that if $xy \geq 2$ we must get that: $$2(b+2) \leq (b+2)xy < 2x^2$$This implies that we must have that: $$b+2 <x^2$$but this must hold true since $x=2a+y$ and $b \leq 2a$. Thus we just need to cover the case when $xy=1$, which implies $x=y=1$, but this means that $a=0$, but $a>0$, thus contradiction.

dame dame

That's a marvellous problem. It's a pity that it did not make it to the IMO paper.

test20 wrote: That's a marvellous problem. It's a pity that it did not make it to the IMO paper. I know right?

I love Vieta Jumping For the sake of contradiction, let \begin{align*}a^2+\left\lceil\frac{4a^2}{b}\right\rceil=(a+k)^2, \quad k\in \mathbb{N}\\ \implies \left\lceil\frac{4a^2}{b}\right\rceil=(a+k)^2-a^2=k^2+2ak. \end{align*}Then, rewrite it as an inequality: $$k^2+2ak-1<\frac{4a^2}{b}\ \le k^2+2ak=k(2a+k).$$Rearranging this, we get $$0 \le bk^2+2abk-4a^2<b.$$Now, let $$bk^2+2abk-4a^2=c, \quad 0 \le c <b \ldots (\dagger)$$and consider the following equation: \begin{align*} &4a^2-2abk-bk^2+c=0\\ \iff &(2a+k)^2-(bk+2k)(2a+k)+k^2+c=0. \end{align*}If a solution exists for a given integer $c$, consider a solution with a minimum sum $a+b+k$, and we'll proceed by Vieta Jumping. Let function $$f(t)=t^2-(bk+2k)t+k^2+c$$and the roots of the equation $f(t)=0$ are $2a+k$ and $d$. Using Vieta's formula, we get $$d=bk+2k-2a-k=\frac{k^2+c}{2a+k} \implies d \in \mathbb{N}.$$So, we have a Vieta jump $$(a, b, k) \longmapsto \left(\frac{d-k}{2}, b, k\right)$$which gives $d \ge 2a+k$. Then, we can easily see that $f(k)>0$. [asy][asy] import graph; size(7cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; /* point style */ real xmin = -12.043471074380172, xmax = 13.278842975206606, ymin = -5.4654695717505675, ymax = 9.675026296018025; pen ffvvqq = rgb(1,0.3333333333333333,0); draw((xmin, 0*xmin + 0)--(xmax, 0*xmax + 0), linewidth(0.8)); real f1 (real x) {return 0.2*x^(2)-5;} draw(graph(f1,-12.033471074380172,13.268842975206606), linewidth(0.8) + ffvvqq); label("$2a+k$",(-5,1.4),SE*labelscalefactor); label("$d$",(5,-0.2),SE*labelscalefactor); label("$k$",(-8.7,-0.4),SE*labelscalefactor); draw((-8.28,8.71168)--(-8.28,0), linewidth(0.8) + linetype("4 4")); label("$f(t)$",(8.25,8),SE*labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] \begin{align*} &f(k)= k^2-(bk+2k)k+k^2+c>0\\ &\iff bk^2<c \end{align*}However, $(\dagger)$ gives $bk^2>c$, contradiction. Hence, $a^2+\left\lceil\frac{4a^2}{b}\right\rceil$ is not a square. $\blacksquare$

AFTSOC that we may write $a^2 + (4a^2+r)/b = (a+x)^2$ where $x>0$ and $0\leq r<b$. Then, \[4a^2+r = b(2ax+x^2) = bx(2a+x).\]Let $y=2a+x$, then this becomes \[(y-x)^2 + r= bxy\]We claim that this Diophantine does not have a solution over positive integers (we have $y=2a+x > x > 0$). First, we claim that if $y>x$ then $y^2-x^2<b$ is impossible. Firstly, $x=y$ will never be a solution, and thus \[bxy \geq 2b = b+b > (y-x)(y+x) +b > (y-x)^2 + r \]a contradiction. Now, assume $y^2-x^2\geq b$. We now Vieta jump. Assume $y>x$, then \[y^2-(bx+2x)y + (x^2+r)\]then we may construct a new solution \[(y,x) \mapsto (y',x) = (bx+2x-y,x) = (\frac{x^2+r}{y},x)\]Note that $b,x,2$ are all integers, so $y'$ is also an integer. Additionally, \[0\leq y' = \frac{x^2+r}{y} \leq \frac{x^2+b-1}{y}\leq \frac{x^2+y^2-b+b-1}{y} < y.\]Thus, we will always be able to descend, where $x+y$ is consistently decreasing and is always an integer, a contradiction so we're done. $\blacksquare$.

Suppose that $a^2+\left\lceil\frac{4a^2}b\right\rceil$ is a square. Then this is equivalent to $a^2+\frac{4a^2+r}b=c^2$ for $0\leq r\leq b-1$, so $a^2b+4a^2+r=bc^2$. Let $a=\frac{x-y}2$ and $c=\frac{x+y}2$. Then, we must have $$b(a-c)(a+c)+4a^2+r=0,$$so $$-bxy+x^2-2xy+y^2+r=0.$$Let $b$ and $r$ be fixed. Consider the solution $(x,y)$ such that $x+y$ is minimal. Then, assume without loss of generality $x\geq y$. Then, since this is a quadratic in $x$, we have that $((b+2)y-x,y)=\left(\frac{y^2+r}x,y\right)$ is also a solution. This is a pair of positive integers since $(b+2)y-x$ is an integer and $\frac{y^2+r}x$ is positive. Therefore, we must have $(b+2)y-x\geq x$, so $(b+2)y\geq2x\geq2y$. However, this is a contradiction since $b$ is a positive integer unless $x=y=0$. However, this means that $a=0$, which is also a contradiction. Therefore, it is impossible for $a^2+\left\lceil\frac{4a^2}b\right\rceil$ to be a perfect square.

I'm thinking there's no way it's this easy. Suppose $a^2+\lceil{\frac{4a^2}{b}}\rceil=c^2$ then $(c+a)(c-a)=\lceil{\frac{4a^2}{b}}\rceil.$ Let $\frac{4a^2+d}{b}$ be $\lceil{\frac{4a^2}{b}}\rceil.$ Let $m=c+a$ and $n=c-a$ then we have $bmn=(m-n)^2+d.$ Now, $m^2-(2n+bn)m+n^2+d$ gives two solutions for $m$: $m$ and $\frac{n^2+d}{m}.$ We claim that $n^2+d<m^2.$ This is equivalent to showing that $4ac>d.$ Note that $d<b.$ If $4a^2<b$ then the ceiling is $1$ which obviously gives no solution $c.$ Otherwise, $4ac>4a^2\ge b>d$ as desired. The fact that we can always find a smaller solution is a contradiction, so we're done.

Assume $(a+n)^2 = a^2 + \left\lceil\frac{4a^2}b\right\rceil$. Part 1: Conversion into algebra \begin{align*} &\iff 2an + n^2 = \left\lceil\frac{4a^2}b\right\rceil \\ &\iff 2an + n^2 \geq \frac{4a^2}b> 2an + n^2 - 1 \\ &\iff \frac{4a^2}{2an + n^2 - 1} > b \ge \frac{4a^2}{2an + n^2}. \tag{1} \\ \end{align*} Claim 2: $\frac{4a^2}{2an + n^2} = \frac{2a}{n} - \frac{2a}{2a + n}$. Proof: \begin{align*} \frac{2a}{n} - \frac{2a}{2a + n} &= \frac{4a^2 + 2an}{n(2a+n)} - \frac{2an}{n(2a+n)} \\ &= \frac{4a^2 + 2an - 2an}{n(2a + n)} = \frac{4a^2}{2an + n^2}. \quad \square \\ \end{align*} Claim 3: $\frac{4a^2}{2an + n^2 - 1} = \frac{2a}{n} - \frac{2a}{n} \left( \frac{n^2 - 1}{2an + n^2 - 1} \right)$. Proof: \begin{align*} \frac{2a}{n} \left (1 - \frac{n^2 - 1}{2an + n^2 - 1} \right) &= \frac{2a}{n} \left( \frac{2an + (n^2 - 1) - (n^2 - 1)}{2an + n^2 - 1} \right) \\ &= \frac{2a}{n} \left( \frac{2an}{2an + n^2 - 1} \right) = \frac{4a^2}{2an + n^2 - 1}. \quad \square \\ \end{align*} Claim 4: $n \le 2a$. Proof: $\left\lceil\frac{4a^2}b\right\rceil \le 4a^2$ for $b \in \mathbb{N}$. Hence, $(a+n)^2 = a^2 + \left\lceil\frac{4a^2}b\right\rceil \le 5a^2 \le 9a^2 \Rightarrow a + n \le 3a \Rightarrow n \le 2a$. $\square$ Let $d = 2a$. By $(1)$ and Claims 2, 3, \begin{align*} \frac{d}{n} - \frac{d}{n}\left( \frac{n^2 - 1}{dn + n^2 - 1} \right) > b \ge \frac{d}{n} - \frac{d}{d+n} \tag{5} \end{align*} By Claim 4, we can let $d = kn - c$, for $k \in \mathbb{N} \ge 2$, $0 \le c < n$. Now $(5)$ becomes: \begin{align*} k - \frac{c}{n} - \frac{kn - c}{n} \left( \frac{n^2 - 1}{kn^2 - cn + n^2 - 1} \right) &> b \ge k - \frac{c}{n} - \frac{kn-c}{(k+1)n - c} \\ \Rightarrow - \frac{c}{n} - \frac{kn - c}{n} \left( \frac{n^2 - 1}{kn^2 - cn + n^2 - 1} \right) &> b - k \ge - \frac{c}{n} - \frac{kn-c}{(k+1)n - c} \tag{6} \end{align*} $0 \le c < n \Rightarrow 0 \le \frac{c}{n} < 1$, and furthermore $kn-c < (k+1)n-c$ so $0 \le \frac{kn-c}{(k+1)n-c} < 1$. Thus, $-2 < RHS \le 0$. Similarly, $n^2 \ge 1 \Rightarrow \frac{n^2 - 1}{kn^2 - cn + n^2 - 1} \ge 0$. Thus, $LHS \le 0$, but $LHS > RHS$ so $LHS > -2$ as well. Part 2: Fixing bounds between integers Now, it suffices to show that \begin{align*} \frac{kn-c}{n} \left( \frac{n^2 - 1}{(k+1)n^2 - cn - 1} \right) - \frac{n-c}{n}&\ge 0 \tag{7} \\ \iff \frac{kn-c}{(k+1)n - c} - \frac{n-c}n &\ge 0 \tag{8} \end{align*}and that neither expression equals zero, for $n \in \mathbb{Z}$, $0 \le c < n$, because this implies \begin{align*} - \frac{c}{n} - \frac{kn - c}{n} \left( \frac{n^2 - 1}{(k+1)n^2 -cn - 1} \right) &\le -1 \\ \iff - \frac{kn - c}{n} \left( \frac{n^2 - 1}{(k+1)n^2 -cn - 1} \right) &\le - \frac{n-c}{n} \\ \iff \frac{kn - c}{n} \left( \frac{n^2 - 1}{(k+1)n^2 -cn - 1} \right) &\ge \frac{n-c}{n} \\ \iff \frac{kn-c}{(k+1)n - c} - \frac{n-c}n &\ge 0 \\ \iff - \frac{kn-c}{(k+1)n - c} &\le - \frac{n-c}n \\ \iff - \frac{c}{n} - \frac{kn-c}{(k+1)n - c} &\le -1 \end{align*}so both bounds always lie within either $(-2, -1)$ or $[-1, 0]$, and hence $b-k$ does too. However, note that $0 \ge LHS > b-k$, so $b-k \neq 0$. Furthermore, if either bound is -1, (7) or (8) must be equality, which contradicts our claim. Thus, $b-k$ ends up in $(-2, -1)$ or $(-1, 0)$, so cannot be an integer. But $b-k \in \mathbb{Z} \iff b \in \mathbb{Z}$ since $k$ integer, so $b \not \in \mathbb{Z}$, which is a contradiction. Part 3: Proof of equivalence \begin{align*} (7) \iff (kn-c)(n^2-1) - (n-c)((k+1)n^2 - cn - 1) &\ge 0 \\ \iff kn^3 - cn^2 - kn + c - (k+1)n^3 + c(k+1)n^2 + cn^2 - c^2n + n - c &\ge 0 \\ \iff -kn - n^3 + c(k+1)n^2 - c^2n + n &\ge 0 \\ \iff -n^2 + c(k+1)n + (1 - k - c^2) &\ge 0. \tag{9} \end{align*} \begin{align*} (8) \iff n(kn-c) - ((k+1)n - c)(n-c) &\ge 0 \\ \iff kn^2 - cn - (k+1)n^2 + c(k+1)n + cn - c^2 &\ge 0 \\ \iff -n^2 + c(k+1)n - c^2 &\ge 0. \tag{10} \end{align*} \begin{align*} (9) \iff n &\in \left [ \frac{-c(k+1) + \sqrt{c^2(k+1)^2 + 4(c^2 + k - 1)}}{-2}, \frac{-c(k+1) - \sqrt{c^2(k+1)^2 + 4(c^2 + k - 1)}}{-2} \right] \\ \iff n &\in \left [ \frac{c}{2} \left((k+1) - \sqrt{(k+1)^2 + 4\left (1 + \frac{k-1}{c^2} \right)} \right), \frac{c}{2} \left((k+1) + \sqrt{(k+1)^2 + 4(1 + \frac{k-1}{c^2})} \right) \right]. \end{align*}Note that the lower bound $< \frac{c}{2} ((k+1) - \sqrt{(k+1)^2}) = 0$, so $(9) \iff n \le \frac{c}{2} \left((k+1) + \sqrt{(k+1)^2 + 4(1 + \frac{k-1}{c^2})} \right)$. Similarly, $(10) \iff n \le \frac{c}{2} ((k+1) + \sqrt{(k+1)^2 + 4})$, but because $k \ge 2$, all differences of two squares are $\ge 7$, and hence $(k+1)^2 + 4 < (k+2)^2$. Thus, $(10) \iff n \le \frac{c}{2} ((k+1) + (k+1)) = c(k+1)$. Now \begin{align*} n &> c(k+1) \\ \Rightarrow (k+1)^2 + 4\left (1 + \frac{k-1}{c^2} \right) &\ge (k+3)^2 \\ \Rightarrow 4 + 4 \cdot \frac{k-1}{c^2} &\ge 4k + 8 \\ \Rightarrow \frac{k-1}{c^2} &\ge k + 1 \tag{11} \end{align*}If $c = 0$, (9) and (10) both yield $-n^2 \ge 0$, a contradiction as $n \in \mathbb{Z}$. If $c \ge 1$, (11) is false. Part 4: Proof that equality is impossible For (10), as above, $(k+1)^2 < (k+1)^2 + 4 < (k+2)^2$ and so $\sqrt{(k+1)^2 + 4}$ isn't an integer, and $n$ isn't either. Contradiction. For (11), if $c = 0$, $n = 0$, contradiction. If $c = 1$, $(k+1)^2 < (k+1)^2 + 4k < (k+3)^2$, so $(k+1)^2 + 4k = (k+2)^2 \Rightarrow 2k = 3$, also contradiction. If $c \ge 2$, $(k+1)^2 < (k+1)^2 + 4\left (1 + \frac{k-1}{c^2} \right) < (k+1)^2 + k < (k+2)^2$, so once again $n$ isn't an integer.

Assume the contrary then $a^2+\tfrac{4a^2+k}{b}=n^2$ with $b>k.$ Set $n=\tfrac{x+y}{2}$ and $a=\tfrac{x-y}{2}$ thus $x^2+y^2-$ $bxy-2xy+k=0.$ Fix $b$ and $k$ and take positive integer solutions $(x,y)$ such that $x+y$ is minimal. WLOG $x\geq y.$ By vieta jumping we have another solution $(z,y)=$ $(\tfrac{y^2+k}{x},y)=$ $((b+2)y-x,y).$ Note that $z$ is also a positive integer. Note that $x^2-y^2>(x-y)^2.$ But we have $b>k=bxy-(x-y)^2$ which implies $(x-y)^2>b>k.$ So $x>\tfrac{y^2+k}{x}= z,$ a contradiction with minimality.

Assume the opposite, so $a^2+\frac{4a^2+\epsilon}{b}=c^2$ for $c\in \mathbb{Z}^+,\epsilon \in [0;b)$. Let $m=c-a,n=c+a$, so $$(n-m)^2+\epsilon =bmn\implies n^2-(2+b)mn+(m^2+\epsilon )=0 \text{ } (\bigstar)$$Note that $n_0=(2+b)-n=\frac{m^2+\epsilon}{n}\in \mathbb{Z}^+,$ so minimizing equality $\bigstar$ by $n$ we obtain $$n\leq \frac{m^2+\epsilon}{n}\implies 4a^2<4ac=n^2-m^2\leq \epsilon<b.$$Therefore $c^2=a^2+1,$ which is absurd, thus the contradiction.

We shall show the related expression $a^2+4\left\lceil\frac{a^2}b\right\rceil$ can never be a perfect square. Suppose otherwise, and consider the smallest $a$ for which such an expression is a perfect square, and say $a^2+4\left\lceil\frac{a^2}b\right\rceil=(a+y)^2$. Then $4\left\lceil\frac{a^2}b\right\rceil = 2ay + y^2$. Clearly $y$ is even, so letting $y=2x$ we get that $\left \lceil \frac{a^2}{b} \right \rceil = ax+x^2$ Claim: $b= \left \lfloor \frac{a}{x} \right \rfloor$ Proof: We have the inequalities $b(ax+x^2-1) < 4a^2 \le b(ax+x^2)$ Observe that \[\frac{a}{x} \cdot (ax+x^2-1) = a^2+ax - \frac{a}{x} \ge a^2\]implying $b \le \frac{a}{x}$. On the other hand, \[(\frac{a}{x} -1)(ax+x^2) = a^2 - x^2 < a^2 \implies b > \frac{a}{x}-1\]Since $b$ is an integer, $b= \left \lfloor \frac{a}{x} \right \rfloor$ as desired. Next, let $a=bx+r$. Then \[\left\lceil\frac{a^2}b\right\rceil = ax+x^2 \implies \left\lceil\frac{b^2x^2+2bxr+r^2}{b} \right\rceil = (bx+r)x+x^2 \implies bx^2+2rx+ \left \lceil \frac{r^2}{b} \right \rceil = bx^2 + rx + x^2 \implies x^2-rx-\left \lceil \frac{r^2}{b} \right \rceil=0 \] For the equation to have integer solutions in $x$, the discriminant of this quadratic must be a perfect square, in other words $r^2+4\left\lceil \frac{r^2}{b} \right \rceil$ is a perfect square. But $r<a$, contradicting the minimality of $a$! Thus $a^2+4\left\lceil\frac{a^2}b\right\rceil$ is never a perfect square. To finish, observe that if $a^2+\left \lceil \frac{4a^2}{b} \right \rceil$ is a perfect square then $4a^2 + 4 \left \lceil \frac{4a^2}{b} \right \rceil = (2a)^2 + 4 \left \lceil \frac{(2a)^2}{b} \right \rceil$ is a perfect square, contradicting the earlier result.

Suppose it equals $(a+k)^2$ ($k$ must be positive integer). First, I claim that $b = \left \lfloor \dfrac{2a}{k}\right \rfloor$. Note that the problem condition is equivalent to \[-1< \frac{4a^2}{b}-k^2-2ak \leqslant 0\]\[\iff \frac{4a^2}{k^2+2ak} \leqslant b < \frac{4a^2}{k^2+2ak-1} \ \ \ \ \ \ (\clubsuit)\]But $\dfrac{4a^2}{k^2+2ak-1} \le \dfrac{2a}{k}$, so $b\leqslant \lfloor \dfrac{2a}{k}\rfloor$. Now it suffices to prove that $b>\dfrac{2a}{k}-1$. Using the left inequality of $(\clubsuit)$, \[b-\frac{2a}{k}\ge\frac{4a^2}{k^2+2ak} - \frac{4a^2}{2ak} = -\frac{4a^2k^2}{2ak(k^2+2ak)}>-1.\]This concludes the proof of the claim. Now write the euclidean division of $2a$ by $k$: \[2a = kb+r,\ 0\le r<k\]\[\implies k^2+2ak = \left\lceil \frac{(kb+r)^2}{b} \right \rceil =\left\lceil \frac{r^2}{b} \right \rceil+k^2b+2kr \]\[\implies k^2+(kb+r)k = \left\lceil \frac{r^2}{b} \right \rceil+k^2b+2kr \]\[\implies k^2-kr =\left\lceil \frac{r^2}{b} \right \rceil\]So in particular $r$ is positive. Now looking at the discriminant (as a quadratic on $k$), we get that \[r^2+4\left\lceil \frac{r^2}{b} \right \rceil = (r+2l)^2.\]Let $x>0$ be minimum such that $x^2+4\left\lceil \frac{x^2}{b} \right \rceil$ is a square. \[\implies x^2+4\left\lceil \frac{x^2}{b} \right \rceil = (x+2y)^2\] with $y>0$. Hence, \[\left\lceil \frac{x^2}{b} \right \rceil = xy+y^2\]I claim $b = \lfloor\frac{x}{y}\rfloor$. \[xy+y^2-1 < \frac{x^2}{b} \leqslant xy+y^2\]\[\frac{x^2}{xy+y^2}\leqslant b < \frac{x^2}{xy+y^2-1}\]\[\frac{x}{y}-1 <\frac{x^2}{xy+y^2}\leqslant b < \frac{x^2}{xy+y^2-1}\leqslant \frac{x}{y} \ \ \ \square\] So we can divide $x$ by $y$: \[x = by+c.\] \[\left\lceil \frac{(by+c)^2}{b} \right \rceil = (by+c)y+y^2\]\[\left\lceil \frac{c^2}{b} \right \rceil = y^2-cy\]by the quadratic formula, \[c^2-4\left\lceil \frac{c^2}{b} \right \rceil\]is a square, a contradiction if $x>c>0$. But $b,y>0 \implies c<x$ so $c=0$. \[\implies xy = \left\lceil \frac{x^2}{b} \right \rceil = xy+y^2\] a contradiction.

I groupsolved this many years ago but here I am again. Suppose that it does equal a square, i.e. $$a^2+\left\lceil \frac{4a^2}{b}\right\rceil=c^2 \implies \left\lceil \frac{4a^2}{b}\right\rceil=c^2-a^2 \implies b(c^2-a^2-1)<4a^2\leq b(c^2-a^2).$$Motivated by the many squares, we perform the classic substitution $c+a= x$ and $c-a= y$, so $a=\tfrac{x-y}{2}$ and the inequality miraculously becomes $$bxy-b<(x-y)^2\leq bxy.$$Thus $(x-y)^2=bxy-d \implies x^2-(b+2)xy+y^2+d=0$ where $0 \leq d < b$. We now apply the technique of Vieta jumping. For fixed $b$ and $d$, suppose that there exist solutions in $\mathbb{Z}^+$ to the equation. Consider the pair $(x_1,y)$ of positive integers such that $x_1+y$ is minimal, and WLOG let $x_1>y$ ($x=y$ means $0>bxy-b\geq 0$: absurd). Viewing the equation as a quadratic in $x$ with $y$ fixed, we have two roots: one of which is $x_1$ and the other $x_2$. Since $x_1+x_2$ and $x_1x_2$ are both positive integers, $x_2$ is a positive integer as well, so by minimality $x_2\geq x_1>y$. Now we have $$x_1+x_2=(b+2)y \text{ and } x_1x_2=y^2+d<y^2+b.$$Since $x_1x_2$ increases if $x_1+x_2$ are fixed and $x_1$ and $x_2$ approach each other, from the first equation and the fact that $x_1\geq y$, $$x_1x_2\geq (b+1)y^2,$$so $y^2+b>(b+1)y^2 \implies b>by^2$, which is absurd. Hence no solutions can exist, implying that the expression given in the problem statement can never be a perfect square. $\blacksquare$

Suppose that \[a^2+\left\lceil\frac{4a^2}b\right\rceil\]is a perfect square for some choice of $a$ and $b$. Then, this is \[ a^2 + \frac{4a^2+m}{b} = k^2 \]for some $0 \le m < b$. This can be rewritten as \[ (b+4)a^2 - bk^2 = -m. \]Using the well-known substitution $k = \frac{x+y}{2}$ and $a = \frac{x-y}{2}$, this becomes a Diophantine equation with the LHS a quadratic form in $x$ and $y$: \[ x^2 - (b+2)xy + y^2 = -m. \]It suffices to prove that this has no solutions for $x \ge y \ge 1$, which can be shown using Vieta jumping. Define the transform $T$ so that \[ T(x,y) =\left( y, \frac{y^2+m}{x} \right) = (y, (b+2)y-x)\]for $x, y \ge 1$; note that $T(x, y)$ is a solution to $x^2 - (b+2)xy + y^2 = -m$ if $(x, y)$ is a solution. If $xy \ge 2$, then \[ (x-y)^2 > bxy-m > bxy-b \ge b, \]from which \[ x^2-y^2 = (x-y)(x+y) > (x-y)^2 > b > m. \]Thus, $0 < \frac{y^2+m}{x} < x$ for all $(x, y)$ with $xy \ge 2$. Now, for a given solution $(x, y)$ with $xy \ge 2$, repeatedly apply $T$; each time, a valid solution $(x_n, y_n)$ with $x_n, y_n \ge 1$ and $\max(x_n, y_n)<\max(x_{n-1}, y_{n-1})$ is generated. Clearly an infinitude of solutions is obtained by this. However, the maximum of the numbers in the pair eventually hits $1$, after which no solution can be generated. Thus no $(x, y)$ with $xy \ge 2$ works. Hence, $(x, y)=(1, 1)$ is the only possible solution, but this implies $b=m$, so it doesn't work. All in all, there are no solutions $(a, b)$ to the original Diophantine equation, as desired.

Suppose it was a perfect square for some $a,b$. Let $\frac{4a^2 + c}{b} = \left \lceil \frac{4a^2}{b} \right\rceil $, where $0\le c< b$. The equation becomes \[a^2 + \frac{4a^2 + c}{b} = d^2,\]where $d$ is a positive integer. Multiplying both sides by $b$ gives $(b+4)a^2 - b d^2 = -c$. Obviously $d>a$, so let $x = d+a$, and $y = d-a$, so that $d = \frac{x+y}{2}$ and $a = \frac{x-y}{2}$ and $x,y$ are positive integers. After multiplying both sides by $4$, the equation becomes $(b+4)(x-y)^2 - b (x+y)^2 = -4c$, so \[ x^2 - (b+2)xy + y^2 = -c\] Notice that if $(x,y)$ is a solution (where $x>y$), then we get another solution \[((b+2)y -x, y) = \left( \frac{y^2 + c}{x}, y \right) \] Claim: Either $(x,y) = (1,1)$ or $c< x^2 - y^2$. Proof: Suppose $(x,y)\ne (1,1)$, so $xy\ge 2$. We have \begin{align*} x^2 - y^2 = (x-y)(x+y) > (x-y)^2 \\ = bxy - c > bxy - b \\ \ge b > c, \\ \end{align*}as desired. $\square$ Thus, if $xy>1$, $0< \frac{y^2 + c}{x} < x$ which means we have another solution with lower maximum value. Thus we can keep repeating such an operation until we get $(1,1)$ is a solution. However, now $2 - (b+2) = -c$, which means $b = c$, which we know isn't true. Therefore there are no solutions.

nice Suppose for the sake of contradiction the expression can take on square values. This is equivalent to \[a^2+\frac{4a^2+c}{b} = k^2,\]where $k$ is a positive integer, and $c$ takes on any value $0 \leq c <b.$ Rewrite this as $(b+4)a^2-bk^2=-c.$ Obviously, $k>a,$ so we use the clever substitution $a=\tfrac{x-y}{2}$ and $k=\tfrac{x+y}{2},$ where $x$ and $y$ take on positive integer values. Slogging through algebra: \[(b+4)\left(\frac{(x-y)^2}{4}\right) - b\left(\frac{(x+y)^2}{4}\right) = -c \implies x^2-(b+2)xy+y^2=-c.\]We now Vieta Jump \[(x,y) \to \left(y, \frac{y^2+c}{x}\right) = (y, (b+2)y-x).\]Suppose $(x,y) \neq (1,1).$ Then we prove $\tfrac{y^2+c}{x}<x.$ Observe that $xy \geq 2.$ From our quadratic in $x,$ we know \[(x-y)^2-bxy=-c>-b \implies (x-y)^2>bxy-b \geq b > c.\]But \[(x-y)^2<(x-y)(x+y)=x^2-y^2,\]implying $c<x^2-y^2.$ Therefore, if minimal $(x,y)$ were to exist, then our process terminates only when we reach the base pair $(x,y)=(1,1).$ Otherwise, using our transformation, we would be able to find a smaller pair. But when plugged back into our original equation, this yields $2-(b+2)=-c \implies b=c,$ which is absurd.

Wow this solution is identical to #37 and #38. !!!! (Also 2019 ISL N8) Let $a$ and $b$ be two positive integers. Prove that the integer \[a^2+\left\lceil \frac{4a^2}{b}\right\rceil\]is not a square. Wow, this is really interesting. There are no actual theorems used here, just some clever observations and a bit of luck. Lemma: The quantity \[a^2+4\left\lceil \frac{a^2}{b}\right\rceil\]is not a perfect square. Proof: Suppose otherwise. Take $k\ge 1$. Write \[a^2+4\left\lceil \frac{a^2}{b}\right\rceil=(a+2k)^2\implies \left\lceil \frac{a^2}{b}\right\rceil = ak+k^2.\]Now comes the first clever observation. If $b\ge \frac{a}{k}$, then \[\frac{a^2}{b}\le ak\implies \left\lceil \frac{a^2}{b}\right\rceil\le ak\]which can't happen. Likewise if $b\le \frac{a-k}{k}$ then \[\left\lceil \frac{a^2}{b}\right\rceil\ge \frac{a^2}{b}\ge \frac{a^2k}{a-k}>ak+k^2\]hence $b\in \left(\frac{a}{k}-1,\frac{a}{k}\right)$. As a result $k\nmid a$ and \[b=\left\lfloor \frac{a}{k}\right\rfloor.\]Write $a=bk+c$ with $1\le c<k$. Then \[\left\lceil \frac{a^2}{b}\right\rceil=ak+k^2\iff \left\lceil \frac{b^2k^2+2bkc+c^2}{b}\right\rceil=bk^2+kc+k^2\]so that \[bk^2+2kc+\left\lceil \frac{c^2}{b}\right\rceil=bk^2+kc+k^2\implies k^2-kc-\left\lceil \frac{c^2}{b}\right\rceil=0.\]Well, here's nice observation #2: since we have a quadratic with integer coefficients with an integer $k$ as the root, it follows that the discriminant \[c^2+4\left\lceil \frac{c^2}{b}\right\rceil\]is a square. But wait! This is analogous to the original expression with $a\to c$ (it's important that $c\ge 1$). If we prove that $c<a$, we should have a contradiction (classic infinite descent argument). But now, $c<k$ and so we need to prove $k\le a$, which is obvious from \[a^2\ge \left\lceil \frac{a^2}{b}\right\rceil=ak+k^2.\]The lemma is proven. It might seem strange that we just moved the $4$ outside, but it turns out the case where $4$ is inside is analogous. Write \[a^2+\left\lceil \frac{4a^2}{b}\right\rceil=(a+k)^2=a^2+2ak+k^2\]and now \[\left\lceil \frac{4a^2}{b}\right\rceil=2ak+k^2\]so we've just taken $a\to 2a$ and may proceed with the same strategy as used in the lemma. Done!

My Vieta Jump is a bit different than the ones posted, so hopefully I am not wrong. (and if irs right, then yay my first N8! ) Assume not. Call $(a,b)$ good if it satisfies the above equation. Extend the domain of $a$ to non-zero integers. See that $(a,b)$ is good if and only if $(-a,b)$ is good. Let $(a,b)=(c,d)$ be the minimal solution which works with $c>0$ (so in other words $c+d$ is minimal). Let \[X^2=c^2+\left\lceil\frac{4c^2}d\right\rceil \iff d(X^2-c^2)=4c^2+r \text{ for some $0 \leq r < d$}\]\[\iff 4c^2-2cdk+r-dk^2=0\]with $X=c+k$ with $k>0$ obviously. Let \[-c'=\frac{r-dk^2}{4c}=\frac{dk}2-c\]now see that $c'$ satisfies $X^2={(c')}^2+\left\lceil\frac{4{(c')}^2}d\right\rceil $ and since $c'$ is obviously rational, it must be an integer. And it is positive as well as $r<d \leq dk^2$. And so $(c',d)$ is also good with $c \in \mathbb{N}$ and hence we get \[c \leq c' = \frac{dk^2-r}{4c} \iff 4c^2 \leq dk^2-r \leq dk^2\]\[\implies \frac{4c^2}d \leq k^2 \iff \left\lceil\frac{4c^2}d\right\rceil \leq k^2\]\[\implies X^2={(c+k)}^2 \leq c^2+k^2\]which is a clear contradiction, as desired.

ihategeo_1969 wrote: My Vieta Jump is a bit different than the ones posted, so hopefully I am not wrong. (and if irs right, then yay my first N8! ) Assume not. Call $(a,b)$ good if it satisfies the above equation. Extend the domain of $a$ to non-zero integers. See that $(a,b)$ is good if and only if $(-a,b)$ is good. Let $(a,b)=(c,d)$ be the minimal solution which works with $c>0$ (so in other words $c+d$ is minimal). Let \[X^2=c^2+\left\lceil\frac{4c^2}d\right\rceil \iff d(X^2-c^2)=4c^2+r \text{ for some $0 \leq r < d$}\]\[\iff 4c^2-2cdk+r-dk^2=0\]with $X=c+k$ with $k>0$ obviously. Let \[-c'=\frac{r-dk^2}{4c}=\frac{dk}2-c\]now see that $c'$ satisfies $X^2={(c')}^2+\left\lceil\frac{4{(c')}^2}d\right\rceil $ and since $c'$ is obviously rational, it must be an integer. And it is positive as well as $r<d \leq dk^2$. And so $(c',d)$ is also good with $c \in \mathbb{N}$ and hence we get \[c \leq c' = \frac{dk^2-r}{4c} \iff 4c^2 \leq dk^2-r \leq dk^2\]\[\implies \frac{4c^2}d \leq k^2 \iff \left\lceil\frac{4c^2}d\right\rceil \leq k^2\]\[\implies X^2={(c+k)}^2 \leq c^2+k^2\]which is a clear contradiction, as desired.

this is like the most canonical Vieta jumping problem I've ever seen Let $\frac{4a^2+r}b \in \mathbb Z$ with $0 \leq r < p-1$. The condition is equivalent to writing \[a^2b+4a^2+r = k^2b.\]Using the substitution $(a, k) = \left(\frac{m-n}2, \frac{m+n}2\right)$ with $m > n$ positive integers, the equation rewrites as \[m^2+n^2-(b+2)mn+r=0.\]Consider the solution $(m_0, n_0)$ to this equation with $m_0+n_0$ minimal. As both $\left(\frac{n_0^2+r}{m_0}, n_0\right)$ is also a solution, it follows that $m_0^2 - n_0^2 \leq r$, implying $m_0 \leq \frac{r+1}2 < \frac{b+1}2$. In particular, $a \leq \frac b2$, so $\frac{4a^2}b \leq 2a$, i.e. $a^2 + \left \lceil \frac{4a^2}b \right \rceil < (a+1)^2$, contradiction.