Dear Mathlinkers, 1. a = y^2 + m, b = y^ + n 2. ab | a^4 + b^4 + 1 3. gcd(a, b) = 1 4. ab | (a-b)^4 + 1 => ab | (m-n)^4 + 1 5. ((m-n)^4+1)/(y^2+m)(y^2+n) <= 16 (due to first step) Prove : y = 1 Sincerely RodwayWorker @below dj ftw yay

Assume $a,b>1.$ Write $a=m^2+c$ and $b=m^2+d$ for a positive integer $m$ and $0\le c,d\le 2m.$ Since $a\mid b^4+1$ and $b\mid a^4+1$, we have $$(d-c)^4+1\equiv 0\pmod{m^2+c},$$$$(c-d)^4+1\equiv 0\pmod{m^2+d}.$$Remark that $\gcd(a,b)=1,$ so for some positive integer $k,$ $$(c-d)^4+1=k(m^2+c)(m^2+d).$$Notice that $(m^2+c)(m^2+d)\ge m^4$ and $(c-d)^4+1\le 16m^4+1,$ so $k<16.$ $\textbf{Case 1: }$ $p\mid k$ for some odd prime $p.$ We have $$(c-d)^4\equiv -1\pmod{p}\implies\text{ord}_{p}(c-d)\nmid 4,$$$$(c-d)^8\equiv 1\pmod{p}\implies\text{ord}_{p}(c-d)\mid 8.$$Therefore, $\text{ord}_{p}(c-d)=8,$ so by Fermat's Little Theorem, $p\equiv 1\pmod{8}.$ But there are no primes less than $16$ congruent to $1$ modulo $8,$ contradiction. $\textbf{Case 2: }$ $k=8$ or $k=4$ We can manually check that $-1$ is not a fourth power mod $4$ or mod $8,$ so there are no solutions in this case. $\textbf{Case 3: }$ $k=2$ Then, $(c-d)^4+1$ is even, so exactly one of $c,d$ is even. This implies that one of $m^2+c,m^2+d$ is even, so $4\mid(c-d)^4+1,$ contradiction. $\textbf{Case 4: }$ $k=1$ Note that $m^4\le (m^2+c)(m^2+d)<(m+1)^4.$ Therefore, since $(m^2+c)(m^2+d)$ is one more than a fourth power, it must be $m^4+1.$ Now it is easy to check that we must have $m=1,c=1,d=0$ or $m=1,c=0,d=1,$ contradiction. Thus, either $a$ or $b$ is $1,$ so the only solutions are $(1,1),(1,2),(2,1).$

Here is the current draft of the official solution: The only solutions are $(1, 1)$, $(1, 2)$, and $(2, 1)$, which clearly work. Now we show there are no others. Obviously, $\gcd(a,b) = 1$, so the problem conditions imply \[ ab \mid (a-b)^4+1 \]since each of $a$ and $b$ divide the right-hand side. We define \[ k \stackrel{\text{def}}{=} \frac{(b-a)^4 + 1}{ab}. \] Claim: [Size estimate] We must have $k \le 16$. Proof. Let $n = \lfloor \sqrt{a} \rfloor = \lfloor \sqrt{b} \rfloor$, so that $a,b \in [n^2, n^2+2n]$. We have that \begin{align*} ab &\ge n^2(n^2 + 1) \ge n^4 + 1 \\ (b-a)^4+1 &\le (2n)^4+1 = 16n^4+1 \end{align*}which shows $k \le 16$. $\blacksquare$ Claim: [Orders argument] In fact, $k = 1$. Proof. First of all, note that $k$ cannot be even: if it was, then $a$, $b$ have opposite parity, but then $4 \mid (b-a)^4 + 1$, contradiction. Thus $k$ is odd. However, every odd prime divisor of $(b-a)^4 + 1$ is congruent to $1 \pmod{8}$ and is thus at least $17$, so $k = 1$ or $k \ge 17$. It follows that $k = 1$. $\blacksquare$ At this point, we have reduced to solving \[ ab = (b-a)^4+1 \]and we need to prove the claimed solutions are the only ones. Write $b = a+d$, and assume WLOG that $d \ge 0$: then we have $a(a+d) = d^4 + 1$, or \[ a^2 - da - (d^4 + 1) = 0. \]The discriminant $d^2 + 4(d^4 + 1) = 4d^4 + d^2 + 4$ must be a perfect square. The cases $d = 0$ and $d = 1$ lead to pairs $(1, 1)$ and $(1, 2)$. If $d \ge 2$, then we can sandwich \[ (2d^2)^2 < 4d^4 + d^2 + 4 < 4d^4 + 4d^2 + 1 = (2d^2 + 1)^2, \]so the discriminant is not a square. The solution is complete. Remark: [Author remarks on origin] This comes from the problem of the existence of a pair of elliptic curves over ${\mathbb F}_a$, ${\mathbb F}_b$ respectively, such that the number of points on one is the field size of the other. The bound $n^2 \le a,b < (n+1)^2$ is the Hasse bound. The divisibility conditions correspond to asserting that the embedding degree of each curve is $8$, so that they are pairing friendly. In this way, the problem is essentially the key result of https://arxiv.org/pdf/1803.02067.pdf, shown in Proposition 3.

v_Enhance wrote: \begin{align*} ab &\ge n^2(n^2 + 1) \ge n^4 + 1\end{align*} Minor thing: you should also mention the case of $a=b=n^2$.

Let \(n:=\lfloor\sqrt a\rfloor=\lfloor\sqrt b\rfloor\). For \(n=1\), the only pairs that work are \((1,1)\), \((1,2)\), \((2,1)\). We will verify that there are no solutions for \(n\ge2\). Since \(a\ne b\), without loss of generality \(a<b\). Obviously \(\gcd(a,b)=1\), so \(ab\) divides \((b-a)^4+1\). Denote \[k:=\frac{(b-a)^4+1}{ab}.\] Claim: \(k<16\) Proof. Since \(a\ge n^2\) and \(b\le n^2+2n\), but \(\gcd(a,b)=1\), note that \[k=\frac{(b-a)^4+1}{ab}\le\frac{(2n-1)^4+1}{n^2(n^2+1)}<16.\]\(\blacksquare\) Finally, note: If \(2\mid k\), then \((b-a)^4+1\) is even, so \(b-a\) is odd, so \(a\) and \(b\) have opposite parity and \(ab\) is even. Then it follows that \((b-a)^4+1\equiv0\pmod4\), which is impossible. If \(p\mid k\) for some odd prime \(p\), then \(p\) divides \((b-a)^4+1\). It follows that \(\operatorname{ord}_p(b-a)=8\), so \(8\mid\varphi(p)\). This does not hold for any prime \(p<16\). Hence \(k=1\), i.e.\ \((b-a)^4+1=ab\). However \[n^4+1<ab<(n+1)^4+1,\]so no solution exists.

The answer is $(1,1),(1,2),(2,1)$. It is trivial to check that these work. Observe that clearly $a,b$ are coprime. The third condition implies that $a=k^2+m$ and $b=k^2+n$, where $k>0$ and $0 \leq m,n \leq 2k$. This yields: $$k^2+m \mid (k^2+n)^4+1 \implies k^2+m \mid (n-m)^4+1,$$and similarly: $$k^2+n \mid (n-m)^4+1.$$Since $\gcd(k^2+m,k^2+n)=1$ this is sufficient to show that: $$(k^2+m)(k^2+n) \mid (n-m)^4+1.$$However, $|n-m|\leq 2k$, which yields the following inequality: $$\frac{(n-m)^4+1}{(k^2+m)(k^2+n)} \leq 16.$$Now observe: If $n-m$ is even, then the numerator is odd and therefore the LHS is odd (since it must be an integer). If $n-m$ is odd, then the numerator is $2 \pmod{16}$. Since $k^2+m$ and $k^2+n$ are then of different parities, one must be even and therefore the LHS is odd (since it must be an integer). Hence: $$\frac{(n-m)^4+1}{(k^2+m)(k^2+n)} \in \{1,3,\ldots,15\}.$$I now claim that, for any integer $a$, there are no odd primes less than $16$ which divide $a^4+1$. Suppose otherwise and let $p \mid a^4+1$, so: $$a^4 \equiv -1 \pmod{p}.$$Squaring both sides of this yields $a^8 \equiv 1 \pmod{p}$, from which one concludes that $\text{ord}_{p}(a)=8$. Noting that $\text{ord}_p(a) \mid \phi(p)$, one can proceed to test every odd prime in the range $[0,15]$ and find that none of them have a totient which is divisible by $8$. Hence such a prime $p$ does not exist. Putting this with the fact that the aforementioned fraction is odd yields the fact that all of the prime divisors of $$\frac{(n-m)^4+1}{(k^2+m)(k^2+n)}$$are at least $17$. But in fact, if this fraction has any prime factors which are at least $17$, it clearly cannot be in the set $\{1,3,\ldots,15\}$ since it is an integer. Therefore: $$\frac{(n-m)^4+1}{(k^2+m)(k^2+n)}=1 \implies (n-m)^4=(k^2+m)(k^2+n)-1,$$so $(k^2+m)(k^2+n)-1$ must be a perfect fourth power. First note that if $m=n=0$, the equation becomes $k^4-1=0$, which has the solution $k=1$. This generates $(a,b)=(1,1)$. However, if $k \geq 2$, then if $m$ and $n$ are not both $0$: $$k^4<(k^2+m)(k^2+n)-1<(k+1)^4,$$since $k^2+m,k^2+n<(k+1)^2$, so $(k^2+m)(k^2+n)-1$ cannot be a perfect fourth power. Thus, $k=1$. Using the fact that $0 \leq m,n \leq 2k$, it is easy to manually check every possible pair $(m,n)$, which gives $(m,n)=(1,0),(0,1)$. This generates $(a,b)=(1,2),(2,1)$. Having exhausted all possible cases, it follows that the three solutions given are the only ones. $\blacksquare$

Almost the same as above. Note that $ab\mid (a^4+1)(b^4+1)\implies ab\mid a^4+b^4+1\implies ab\mid (a-b)^4+1$. Define $k=\frac{(a-b)^4+1}{ab}$. Define $n=\lfloor a\rfloor=\lfloor b\rfloor$. Then $n^2\le a, b\le n^2+2n$. And $n^4+1\le n^4+n^2=n^2(n^2+1)\le ab\le (b-a)^4+1\le (2n)^4+1=16n^4+1$. Thus $k\le \frac{16n^4+1}{n^4+1}<16$. If $2\mid k$, then $2\mid (a-b)^4+1\implies a\not\equiv b\pmod 2\implies 2\mid ab\implies 4\mid (a-b)^4+1$. However $(a-b)^4\equiv 1\pmod 4$, a contradiction. Therefore $2\not| k$. If $p\mid k$ for some odd prime $p$ then $3\le p\le 13$. Note that $p\mid (a-b)^4+1\implies ord_p(b-a)=8\implies 8\mid \phi(p)=p-1$ which is impossible. Thus $k=1$. Therefore $ab=(a-b)^4+1$. Let $a=n^2+a_1$, $b=n^2+b_1$, where $0\le a_1, b_1\le 2n$. WLOG set $b\ge a$, then $b_1\ge a_1$. For $b_1\ge a_1\ge 1$ or $b_1\ge 2$, $a_1=0$ we have $n^4+1<(n^2+a_1)(n^2+b_1)<(n+1)^4<(n+1)^4+1$. Therefore either $a=b=n^2\implies a=b=1$ or $a=n^2$, $b=n^2+1\implies a=1$, $b=2$. In conclusion, $(a, b)\in \{(1, 1), (1, 2), (2, 1)\}$.

Pretty much the same as above. Just for storage. The solutions are $(1,1),(1,2),$ and $(2,1)$. Note that $\gcd(a,b)=1 \implies ab|(a^4+1)(b^4+1)=a^4b^4+a^4+b^4+1 \implies ab|a^4+b^4+1 \implies ab|(a-b)^4+1$. Let $a=n^2+c$, $b=n^2+d$, where $0 \le c,d \le 2n$, from the floor condition. We have that $(n^2+c)(n^2+d)|(d-c)^4+1$. Let $k=\frac{(d-c)^4+1}{(n^2+c)(n^2+d)}$. We have $k \le \frac{16n^4+1}{(n^2+c)(n^2+d)} \le \frac{16n^4+1}{n^2(n^2+1)}<16$. Note that if $p|k$, then $ord_p(d-c)=8 \implies 8|p-1 \implies p \ge 17$, contradiction. Thus $k=1$. This means $$(n^2+c)(n^2+d)=(d-c)^4+1 \implies (2n^2+c+d)^2-(d-c)^2=4(d-c)^4+4 \implies (2n^2+c+d)^2=4(d-c)^4+(d-c)^2+4$$. If $|d-c|>1$, then $(2(d-c)^2)^2<4(d-c)^4+(d-c)^2+4=(2n^2+c+d)^2<(2(d-c)^2+1)^2$, contradiction. Thus $d-c=-1,0,1$. This means $(a,b) = (1,1)$ or $2n^2+c+d=3 \implies (a,b)=(1,2),(2,1)$.

Pretty much the same as every other solution on this thread but whatever lol. The only solutions are $(a,b) = (1,1), (1,2), (2,1)$. It can be easily verified that these work. Let $n = \lfloor\sqrt{a}\rfloor=\lfloor \sqrt{b}\rfloor$. Claim: $ab \mid (a-b)^4 +1$. Proof. This arises from multiplying the first two problem conditions. We obtain that $$ab \mid a^4b^4 + a^4 + b^4 +1 \ \Longrightarrow \ ab \mid a^4 + b^4 +1 \ \Longrightarrow \ ab \mid (a-b)^4+1 \qquad \square$$This will allow us to bound $a$ and $b$. Set $k = \frac{(a-b)^4+1}{ab}$. Then, we have the following: Claim: $k = 1$. Proof. Note that $$k = \frac{(a-b)^4+1}{ab}\le \frac{(2n)^4+1}{n^4} \le 16$$for $n \ge 2$. Now, consider some prime $p$ that divides $k$. We have that $$(a-b)^4 + 1 \equiv 0 \pmod{p} \quad \Longrightarrow \quad (a-b)^4 \equiv -1 \pmod{p} \quad \Longrightarrow \quad (a-b)^8 \equiv 1 \pmod{p}$$So either $p=2$ or the order of $(a-b) \pmod{p}$ is $8$, implying that $p \equiv 1 \pmod{8}$. As a result, $k$ can only be divisible by $2$ or primes $p \equiv 1 \pmod{8}$, so $k = 1, 2, 4, 8, 16$. Finally, $(a-b)^4+1$ is congruent to $1$ or $2$ $\pmod{8}$, when $a, b$ have the same and opposite parity, respectively. This implies that $\nu_2((a-b)^4+1) = 0, 1$ when $\nu_2(ab)$ is $\ge 0$ or $\ge 1$, respectively. Thus $\nu_2(k) = 0$, so $k=1$. $\square$ Claim: $n = 1$. Proof. Assume FTSOC that $n \ge 2$. We have that $(a-b)^4+1 = ab$. Note that we can bound $$n^4 \le ab \le (n+1)^4 \ \Longrightarrow \ n^4 \le (a-b)^4+1 \le (n+1)^4$$For $n \ge 2$, this implies that $(a-b) = n$, so that $$ab = n^4+1$$However, $$(n^2)^2 < n^4+1 < n^2(n^2+1)$$so $n^4+1$ is between the two smallest consecutive values of the product $ab$, so $ab = n^4+1$ is impossible, contradiction. $\square$ As $n = 1$, it can be easily checked that the aforementioned solutions are indeed the only ones.

We claim the only solutions are $(a,b) = (1,1),(2,1),(1,2)$. Additionally, note that $\gcd(a,b)=1$. Then, \[ab\mid (a-b)^4+1\]For some integer $k$, we have $k^2\leq a,b\leq k^2+2k$, so \[\frac{(a-b)^4+1}{ab}\leq \frac{(2k)^4+1}{(k^2+2k)^2}< 16\]Additionally, $(a-b)^4+1\equiv 0 \pmod{p}$ implies that $p\equiv 1 \pmod{8}$ because $ord_p(a-b)=8$. Thus, the quotient only has factors $p\equiv 1 \pmod{8}$, so the quotient must be 1. Thus, $ab= (a-b)^4+1$. Substituting $b=a+d$ gives \[a^2-d\cdot a -(d^4+1) = 0\]Hence the following discriminant is a square: \[4d^4+d^2+4\]$d=0$ gives $a=1$ and $d=\pm 1$ gives $a=1,2$. For $d\geq 2$, there are no solutions because \[(2d^2)^2 < 4d^4+d^2+4 < (2d^2+1)^2\]thus, we have verified that the original solutions are the only ones possible and we're done. $\blacksquare$.

AwesomeYRY wrote: We claim the only solutions are $(a,b) = (1,1),(2,1),(1,2)$. Additionally, note that $\gcd(a,b)=1$. Then, \[ab\mid (a-b)^4+1\]For some integer $k$, we have $k^2\leq a,b\leq k^2+2k$, so \[\frac{(a-b)^4+1}{ab}\leq \frac{(2k)^4+1}{(k^2+2k)^2}< 16\]Additionally, $(a-b)^4+1\equiv 0 \pmod{p}$ implies that $p\equiv 1 \pmod{8}$ because $ord_p(a-b)=8$. Thus, the quotient only has factors $p\equiv 1 \pmod{8}$, so the quotient must be 1. Thus, $ab= (a-b)^4+1$. Substituting $b=a+d$ gives \[a^2-d\cdot a -(d^4+1) = 0\]Hence the following discriminant is a square: \[4d^4+d^2+4\]$d=0$ gives $a=1$ and $d=\pm 1$ gives $a=1,2$. For $d\geq 2$, there are no solutions because \[(2d^2)^2 < 4d^4+d^2+4 < (2d^2+1)^2\]thus, we have verified that the original solutions are the only ones possible and we're done. $\blacksquare$. You wrote How did you get this part? Additionally, $(a-b)^4+1\equiv 0 \pmod{p}$ implies that $p\equiv 1 \pmod{8}$ because $ord_p(a-b)=8$. Thus, the quotient only has factors $p\equiv 1 \pmod{8}$, so the quotient must be 1. Why must p be 1 mod 8? Am I missing something?

bump for above

$ord_p(x)\mid p-1$ this is because if $x^{p-1}\equiv1\pmod p$ and $x^{ord_p(x)}\equiv1\pmod p$, then $x^{\gcd(p-1,ord_p(x)}\equiv1\pmod p$. Since $ord_p(x)$ is the smallest such number, $ord_p(x)\leq\gcd(p-1,ord_p(x))$, which implies $ord_p(x)\mid p-1$.

1- $mcd(a, b)=1$ 2- $b^4 \equiv -1 \pmod a $ therefore $b^8 \equiv 1 \pmod a$ This means that either $\varphi (a) \vert8$ or $8\vert\varphi(a)$. This is also similar for b. Clearly $8\vert\varphi(a)$ and $8\vert\varphi(b)$ cant be simultaneously true because that would contradict our first result. Therefore $$\varphi(a)\vert8$$$$\varphi(a)\in \{1, 2, 4, 8\}$$If both $\varphi(a)$ and $\varphi(b)$ are greater than 1: $2\vert\varphi(a)$ and $2\vert\varphi(b)$ contradicting our first result. Lets take then $\varphi(a) = 1$. This means that $a \in \{1, 2\}$ trying each case we get all solutions $(1, 2), (2, 1), (1, 1)$. Edit: forgot to mention that $\varphi(a)$ was euler's totient function

WLOG $a<b$ by symmetry. We claim $\gcd(a,b)=1$; if a prime $p\mid a$ and $p\mid b$, then $p\mid b \mid a^4+1$ but also $p\mid a$, so $p\mid 1$, contradiction. Define $d:=b-a$, so $b=a+d$. Note $d>0$. Then $a\mid (a+d)^4+1$, so $a\mid d^4+1$. Also, $a+d\mid a^4+1$, so $a+d\mid d^4+1$. Since $\gcd(a,b)=1$, combining gives $a(a+d)\mid d^4+1$. So now define $k\in \mathbb{N}$ so that \[ d^4+1=ka(a+d). \qquad (\spadesuit) \]Since $\lfloor\sqrt{a}\rfloor=\lfloor \sqrt{b}\rfloor$, so $d \le 2\sqrt{a}$. Then \[ ka^2 < ka(a+d) = d^4+1 \le (2\sqrt{a})^4+1 = 16a^2+1 \implies k<16. \]Claim: $k\equiv 1 \pmod8$. Proof: Suppose a prime $p\mid a$. Then $b^4\equiv -1 \pmod{p}$, so $b^8\equiv 1 \pmod{p}$. So the order of $b\pmod{p}$ divides $8$ and does not divide $4$, so it is in fact $8$. Hence $8\mid p-1$, so $p\equiv 1 \pmod8$. All prime factors of $a$ are $1\pmod8$, so $a\equiv 1\pmod8$. By symmetry, $b\equiv 1 \pmod8$. Hence $d=b-a\equiv 0 \pmod8$. Now taking $(\spadesuit)\pmod8$ gives $k\equiv 1\pmod8$. $\blacksquare$ Therefore, $k\in \{1,9\}$. If $k=9$, then $(\spadesuit)$ becomes $d^4+1=9a(a+d)$. The RHS is $0\pmod3$, but $d^4 \mod 3 \in \{0,1\}$, contradiction. Hence $k=1$, so $(\spadesuit)$ becomes \[ d^4+1=a(a+d). \]Write the above as a quadratic in terms of $a$, then its discriminant is $d^2+4(d^4+1)=(2d^2+1)^2 - 3d^2 + 3$, which must be a square. However, for $d\ge 2$, it is between $(2d^2-1)^2$ and $(2d^2)^2$, contradiction. So $d\le 1$, and the only solutions turn out to be $(a,b) = (1,1),(2,1),(1,2)$.

The only pairs that work is $(1, 1), (1, 2)$, and $(2, 1)$. Claim: $\gcd(a, b)=1.$ Proof: Assume FTSOC that $\gcd(a, b)=x \neq 1,$ so we must have $x \mid a \implies x \mid b^4+1,$ but that is a contradiction because it's implying that $x \mid 1.$ $\Box$ Next, note that we have $$a \mid b^4+1 \implies b^4+1 \equiv 0\pmod{a} \implies (b-a)^4+1 \equiv 0\pmod{a}.$$Similarly, we have $(b-a)^4 +1 \equiv 0\pmod{b}$ so $ab \mid (b-a)^4+1.$ Note that it is easy to see that as long as we have $ab \mid (b-a)^4+1$ $(a, b)$ satisfies the original problem conditions by just assuming $ab \mid (b-a)^4+1$ and working your way backwards with the third condition being true. From the third condition of the problem, let $n=\lfloor\sqrt{a}\rfloor=\lfloor \sqrt{b}\rfloor, a=n^2+a_1, b=n^2+b_1, a_1<2n+1, b_1 < 2n+1.$ Claim: $\frac{(b-a)^4+1}{ab}=1.$ Proof: First assume that $2 \mid \frac{(b-a)^4+1}{ab},$ then consider the parity of $a$ and $b,$ if $a$ and $b$ are the same parity, then $(b-a)^4+1$ is always odd so $k$ can't be even. If $a$ and $b$ are different parity, then $(b-a)^4 \equiv 1 \pmod{4} \implies (b-a)^4+1 \equiv 2 \pmod{4},$ so $v_2((b-a)^4+1)=1,$ but $v_2(ab) \ge 1,$ therefore $k$ can't be even. Now assume an odd prime $p$ divides $k,$ we have $$p \mid k \implies p \mid (b-a)^4+1 \implies p\mid (b-a)^8-1 \implies ord_p(b-a)=8 \implies 8 \mid p-1 \implies p \equiv 1\pmod{8}.$$The least prime that is $1\pmod{8}$ is $17,$ from $a=n^2+a_1, b=n^2+b_1$ we have $$\frac{(b-a)^4+1}{ab} \le \frac{((n^2+2n)-(n^2))^4+1}{(n^2)^2}=\frac{(2n)^4+1}{n^4}=16+\frac{1}{n^4} <17,$$that is true if $n \neq 1.$ If $n=1,$ then the only possible values for $a$ and $b$ are $1, 2, 3$ and it's easy to check that for each of the cases that work, $\frac{(b-a)^4+1}{ab}=1.$ $\Box$ Claim: $n=1.$ Proof: If $n \ge 2,$ then $ab >n^4+1,$ but that is a problem because $$ab-1=(a-b)^4 <(n+1)^4,$$so, we must have $ab=n^4+1$ since $ab-1$ is a 4th power. But that is a contradiction for $n \ge 2.$ $\Box$ So, $n=1$ and there is only $9$ cases to check to get that the only solutions are $(1, 1), (1, 2)$, and $(2, 1)$. $\blacksquare$

Fakesolved

this is legitimately the weirdest problem i have ever solved The answers are $(a,b)=(1,1),(1,2),(2,1)$, all of which obviously work. Clearly $\gcd{(a,b)}=1$ and $ab\mid (a-b)^4+1$. Now by simple bounding we find that \[D^4+1=kP\implies k\mid D^4+1\]where $D=a-b\ge 0$ (WLOG), $P=ab$, and $1\le k\le 16$. The point here is that $D^4+1$ actually misses all of the multiples $\pmod p$ for all primes $p\in \{3,5,7,11,13\}$ (which can be manually verified). In addition, this expression misses all multiples of $4$, so we must have $k\in \{1,2\}$. Now note by the quadratic formula that $D^2+4P$ must be a perfect square; some manipulations give that \[4kD^4+k^2D^2+4k\]must be a perfect square as well. Case 1: $k=1$. In this case, we find that \[t^2-(8D+1)^2=63\]for some $t$. Then we find that $8D+1\in \{1,9,31\}$ so either $D=0$ or $D=1$. These give the solutions $(1,1)$ and $(2,1)$ (with the additional solution of $(1,2)$ as a permutation). Case 2: $k=2$. In this case, by $\pmod 3$ we find that \[4kD^4+k^2D^2\equiv 0\pmod 3\implies 4kD^4+k^2D^2+4k\equiv 2\pmod 3\]which cannot be a square, so this case gives no solutions. We are done. $\blacksquare$

wtfrick is this It is easy to see that $\gcd(a,b)=1$. Then $ab \mid (a-b)^4+1$. Let $\lfloor \sqrt a \rfloor = x$. We have $\frac{(a-b)^4+1}{ab} \leq \frac{(2x)^4+1}{x^4}$, or the value is less than or equal to $16$. We prove this value has to be odd. Since $\gcd(a,b)=1$, at most one is even. If both are odd, it is true. But notice that $(a-b)^4+1 \equiv 2 \pmod 4$ if $a$ is even, $b$ is odd, so the claim is true. Furthermore, note that $x^4\equiv -1$ has no solutions mod $3,5,7,11,13$, and this forces the value to be $1$. To finish, we solve $(a-b)^4+1=ab$. Let $a-b = k$ (wlog, $a>b$. If $a=b$, we get $(1,1)$). We get $k^4 +1 = b(b+k)$. This is a quadratic in $b$, and the discriminant is $k^2 + 4k^4+4$. This is between $(2k^2+1)^2$ and $(2k^2+2)^2$ if $k>1$, so $k=1$. Then $b^2+b-2,$ or $b=1$ and $a=2$. The set of solutions is $(2,1), (1,2), (1,1).$

The only solutions are $(a,b)=(1,1), (2,1), (1,2)$ which can be easily verified. We have $\gcd(a,b)=1$ so we can consider $a\neq b$. Assume WLOG $b>a$, then $$a\mid b^4+1\iff a\mid (b-a)^4+1$$and similarly $b\mid a^4+1$, so $$ab\mid (b-a)^4+1$$Write $a=k^2+x, b=k^2+y$ for $0\leq x,y<2k+1$ so we have that $$(k^2+x)^2+(y-x)(k^2+x)\leq (y-x)^4+1\leq (2k)^4+1<16((k^2+x)^2+(y-x)(k^2+x))$$If $\frac{(b-a)^4+1}{ab}=2m$ for some positive integer $m$, then by letting $n=k^2+x, z=y-x$ we have $2mn(n+z)=z^4+1$ so RHS is odd, so $z$ is even so $2\mid n(n+z)$. Thus, $4\mid LHS$ so $4\mid RHS=z^4+1\equiv 2\pmod{4}$, a contradiction. If this value is odd, notice that $r\mid x^4+1$ has no solutions for $r=3,5,7,9,11,13,15$ so we must have $\frac{(b-a)^4+1}{ab}=1$. Therefore, $n^2+zn-(z^4+1)=0$ which has discriminant $\Delta=4z^4+z^2+4$ which must be a perfect square for $n\in\mathbb Z^+$ to have solutions. But for $z>2$, we have $(2z^2)^2<4z^4+z^2+4<(2z^2+1)^2$ a contradiction, forcing $z=1$. Then, $n=\frac{-1\pm 3}{2}=1$ so $a=1$ and $b=1+1=2$. By symmetry, $a=2,b=1$ is also a solution, completing the proof.

Same strategy to deal with floor as APMO 2013/2: Let $\lfloor a \rfloor = \lfloor b \rfloor = n$ such that $a = n^2+c$ and $b = n^2+d$ for $0 \leq c,d \leq 2n$. First note that $\gcd(a,b) = 1$. This is easy to show by contradiction. If $\gcd(a,b) = g$ for some $g > 1$ let $q$ be a prime factor of $g$. It follows that $q \mid a$ and $q \mid b$. However, if $q \mid a$ then $q \mid b^4+1$ so if $q \mid b$ then $q \mid b^4+1-b^4 = 1$ contradiction. Now plug in our substitution to get $$n^2+c \mid (n^2+d)^4+1$$This is equivalent to $$n^2+c \mid (d-c)^4+1$$Similarly, $$n^2+d \mid (d-c)^4+1$$However we know that $n^2+c = a$ and $n^2+d = b$ are relatively prime so $$(n^2+c)(n^2+d) \mid (d-c)^4+1$$ We will prove that $$\frac{(d-c)^4+1}{(n^2+c)(n^2+d)} < 16$$This is easy by bounding. $$\frac{(d-c)^4+1}{(n^2+c)(n^2+d)} \leq \frac{16n^4+1}{n^2(n^2+1)} < 16$$ Let $\frac{(d-c)^4+1}{(n^2+c)(n^2+d)} = k$. We will prove that in fact $k = 1$. First consider if $2 \mid k$. Note that if $c$ and $d$ are the same parity than $k$ is odd, so $c$ and $d$ need to be opposite parity. If we let $d-c = 2x+1$ then $(d-c)^4+1 \equiv (2x+1)^4+1 \equiv (4x^2+4x+1)^2+1 \equiv 2 \pmod 4$ so $v_2((d-c)^4+1) = 1$. However since one of $c$ and $d$ has to be even we know $v_2((n^2+c)(n^2+d)) \geq 1$ so $2 \nmid k$. If $k$ is divisible by an odd prime $p$, then by Fermat's Christmas Theorem (I see that one can also use orders for the following analysis but whatever) we know $p \equiv 1 \pmod 4$. The only $1 \pmod 4$ primes under $16$ are $5$ and $13$. FLT tells us that $(b-a)^4+1$ cannot be divisible by $5$ and simply bashing all residues mod $13$ for $b-a$ shows that $(b-a)^4+1$ also cannot be divisible by $13$. It follows that no odd primes divide $k$ and that $k$ cannot be even. This means $k = 1$ as desired. If $k = 1$ then $$(d-c)^4 = (n^2+c)(n^2+d)-1$$We first take care of $c = d = 0$ which yields the solution $(1,1)$. Otherwise, we have $$n^4 \leq (n^2+c)(n^2+d)-1 \leq (n^2+2n)^2-1 = (n+1)^4-2(n+1)^2 < (n+1)^4$$ It suffices to then consider $n^4 = (n^2+c)(n^2+d) - 1$ which rearranges to $1-cd = (d+c)n^2$. It follows that $c$ and $d$ can either be $0$ or $1$. Plugging in $(c,d) = (1,1)$ yields an invalid solution. Plugging in $(1,0)$ and $(0,1)$ yield $(2,1)$ and $(1,2)$ respectively. Plugging in $(0,0)$ yields no solutions. We've exhausted all cases and the only solutions are $(1,1), (1,2), (2,1)$.

A standard NT problem

The only solutions are $(1, 1), (1, 2), (2, 1)$, and they clearly work. First, realize that $\gcd(a, b)=1$, so by CRT \[ ab \mid (a-b)^4+1. \]Let \[ k:=\frac{(a-b)^4+1}{ab}. \]The $\lfloor \sqrt{a} \rfloor = \lfloor \sqrt{b} \rfloor$ condition is equivalent to $a, b \in [m^2, m^2+2m]$ for some $m$. Thus we have the bound \[ k \le \frac{(2m)^4+1}{m^2 \cdot m^2} = 16 + \frac{1}{m^4}, \]so $k \le 16$ as it is integral. If $k=1$, then $m^4 \le ab = (a-b)^4+1 < (m+1)^4$, so $ab$ is either $m^4$ or $m^4+1$. In the former case, $a=b=m^2$, but the divisibility condition in the problem implies $m=1$, returning the solution $(1, 1)$. In the latter case, we must have $m^2(m^2+1)=m^4+1$, so $m=1$, returning $(1, 2)$ and $(2, 1)$. Now what follows is an elementary orders argument, based on the primes dividing $k>1$. If $k$ is even, then it follows that $b-a$ is odd, but inspecting modulo $4$ implies that $(b-a)^4 \equiv 3\pmod{4}$, a contradiction. If $k$ is odd and $p$ is an odd prime divisor of $k$, then $(b-a)^4 \equiv -1\pmod{p}$, meaning that $b-a$ has order $8$ modulo $p$. However, this is impossible as none of the primes $p<16$ are $1\pmod{8}$.

By the problem condition($\lfloor{\sqrt{a}\rfloor}=\lfloor{\sqrt{b}}\rfloor$) we have: $$a=k^2+a_{1}$$$$b=k^2+b_{1}$$where $0 \le a_{1},b_{1} \le 2m+1$. (Let $a,b \ge 2$) then the other conditions imply: $$\frac{(a_{1}-b_{1})^4+1}{(k^2+a_{1})(k^2+b_{1})} \in Z $$Since $gcd(k^2+a_{1},k^2+b_{1})=1$. Claim 1: $s=\frac{(a_{1}-b{1})^4+1}{(k^2+a_{1})(k^2+b_{1})} \le 15$ Proof: $s \le \frac{(2m)^4}{m^4}-1=15$ Claim 2:$v_2(s)=0$ Proof: Sps $v_2(s) \ge 1$, then $v_2(ab) \ge 1$, thus $v_2((a_{1}-b_{1})^4+1) \ge 2$ contradiction. Claim3: $v_p(s)=0$, $\forall p \le 15$ Proof: Sps $v_p(s) \ge 1$ for some $p$, then $O_{p}(a_{1}-b{1})|8 $ since $p\ne 2 $ $\implies$ $8| p-1$.contradiction. Claim 4: There exists no solution for $s=1$ Proof: Note that $(k^2+a_{1})(k^2+b_{1})$ is bounded by consecutive squares $k^4$ and $(k+1)^4$ Thus we must have $(k^2+a_{1})(k^2+b_{1})=k^4+1$.Absurd. Thus the only solutions are $(a,b)=(1,2),(2,1),(1,1)$.

Nice and simple Let $\lfloor \sqrt{a} \rfloor = \lfloor \sqrt{b} \rfloor =c$ and $a = c^2 + i, b = c^2 + j$. Then $$c^2 + i | (c^2 + j)^4 + 1 \implies c^2 + i | (i-j)^4 + 1.$$Similarly, $c^2 + j | (i-j)^4 + 1$, and since $$\gcd(c^2 +i, c^2 + j) = \gcd(c^2 + i, i-j) = 1$$we have $(c^2+i)(c^2+j)|(i-j)^4 + 1$. Since $(i-j)^4 + 1 \leq 16c^4$, we can show that there exists $k \leq 15$, such that $k(c^2+i)(c^2+j) = (i-j)^4 + 1$ Note that if $p|(i-j)^4 + 1$ for odd prime $p$, we have $ord_p(i-j) = 8$. Then $8 | \varphi(p)$ impossible for $p<17$. If $2^m | (i-j)^4 + 1$ for $m>1$ then similarly $8 | \varphi(2^m)$, again impossible for $k < 16$. Finally if $2 | (i-j)^4 + 1$, then its easy to see $2 | (c^2+i)(c^2+j)$. This means Nothing can divide $$\frac{(i-j)^4 + 1}{(c^2+i)(c^2+j)},$$so $k=1$ and $(c^2+i)(c^2+j) = (i-j)^4 + 1$. Since $$c^4+1 \leq (c^2+i)(c^2+j) < (c^2 + 2c + 1)^2+1,$$we must have $$(i-j)^4 + 1 = c^4+1 = (c^2+i)(c^2+j)$$which can only happen in $(1, 1), (1, 2), (2,1)$.

Answer: $(a, b) = (1, 1), (2, 1)$. Since $a \mid b^4 + 1$ and $b \mid a^4 + 1$, so $\gcd(a, b) = 1$. WLOG assume $a > b$. Then $b \mid (a - b)^4 + 1$ and $a \mid (a - b)^4 + 1$. So $ab \mid (a - b)^4 + 1$. Let $k = \lfloor \sqrt{a} \rfloor$. Then $a - k^2 \le 2k$ and $b - k^2 \le 2k$, so $a - b \le 2k$. Thus $(a - b)^4 + 1 \le 16k^4 + 1$. So $\frac{(a - b)^4 + 1}{ab} \le \frac{16k^4 + 1}{ab} \le 16$. Now note that if $3 \le p$ prime divides $x^4 + 1$, then $ord_p(2) = 8$, so $8 \mid p - 1$. Thus $\frac{(a - b)^4 + 1}{ab}$ cannot divisible by $p$. Now note that $\nu_2((a - b)^4 + 1) \le 1$ and if $\nu_2((a - b)^4 + 1) = 1$, then $\nu_2(ab) \ge 1$. So $\frac{(a - b)^4 + 1}{ab}$ is odd. Therefore $(a - b)^4 + 1 = ab$. Let $d = (a - b)$, then $b(b + d) = d^4 + 1$, so $b^2 + bd - (d^4 + 1) = 0$. Hence $b = \frac{-d + \sqrt{d^2 + 4d^4 + 1}}{2}$, so $\sqrt{4d^4 + d^2 + 4}$ is integer. If $d > 1$, then $4d^4 < 4d^4 + d^2 + 4 < 4d^4 + 4d^2 + 1$, a contradiction. If $d = 1$, then $b = 1$, hence $a = 2$. If $d = 0$, then $b = 1$. Therefore $(a, b) = (1, 1), (2, 1)$. $\blacksquare$

The answer is $(a,b)=(1,1),(2,1),(1,2)$. If $p$ is a prime that divides both $a$ and $b$, we clearly cannot have $a\mid b^4+1$ since $p\nmid b^4+1$. Hence, $a$ and $b$ are relatively prime. Claim: If $p$ is an odd prime dividing $a$, then $p\equiv 1\pmod{8}$. We have $$p\mid b^4+1.$$This means that the order of $b$ mod $p$ is 8, and thus $p\equiv 1\pmod{8}.$ Clearly a similar result holds for $b$. The main realization here is that the third condition basically says that $a$ and $b$ are "about the same size", which motivates considering $a-b$. We have $$b^4\equiv -1\pmod{a},$$which means that $$(b-a)^4\equiv -1\pmod{a}.$$Let $d=b-a$. Then, this is $$d^4\equiv -1\pmod{a}.$$Similarly we have $$d^4\equiv -1\pmod{b}$$as well, so $$d^4\equiv -1\pmod{ab}$$since $a$ and $b$ are relatively prime. Write this as $$d^4=abk-1.$$Suppose that the floor $\sqrt{a}$ and $\sqrt{b}$ is $n$. Then, we have $$|d|\leq 2n,$$which means that $$d^4\leq 16n^4\leq 16ab.$$Therefore, $k\leq 16.$ Case 1: $a$ and $b$ are both odd. Then, note that $d^4\equiv 0\pmod{8}.$ Furthermore, we know that $a\equiv b\equiv1\pmod{8}$, so taking mod $8$ reveals that $$k\equiv 1\pmod{8}.$$Hence we have either $$(a-b)^4=ab-1$$or $$(a-b)^4=9ab-1.$$The latter can be ruled out by taking mod 3. As for the former, rewrite it as $$d^4=b(b+d)-1$$$$b^2+bd-1-d^4=0.$$Thus, the discriminant, which is $$4d^4+d^2+4,$$has to be a perfect square. However, unless $|d|=0$ or $|d|=1$ (the latter of which ruled out due to $a,b$ odd), this is too close to $(2d^2)^2$ to be a perfect square. If $a=b$, then of course we must have $a=b=1$. Case 2: one of them is even. WLOG $a$ is even. Then, we know that $a\equiv 2\pmod{8}$ and $b\equiv 1\pmod{8}$. Furthermore, $d^4\equiv 1\pmod{8}$. Therefore, taking the equation mod 8 yields $$1\equiv 2k-1\pmod{8}$$$$2k\equiv 2\pmod{8}$$$$k\equiv 1\pmod{4},$$so $k=1,5,9,13.$ We have already done the 1 and 9 cases, but this time $(a,b)=(2,1)$ are allowed in the $k=1$ and $|d|=1$ case as $a$ is now even. Symmetrically, we also have $(1,2).$ The 5 case dies to mod $5$ since $(a-b)^4$ is either 0 or 1 mod 5 but $5ab-1$ is 4 mod 5. Similarly, $-1$ is not a quartic residue mod $13$ so $k=13$ dies to mod $13$. We are done.

This feels like two problems combined into one. Answer: $(a, b) = (1, 1)$, $(2, 1)$, and $(1, 2)$. Claim. [First Main Step] $ab \mid (a-b)^4 + 1$. Proof. We have $a \mid (a-b)^4+1$ as all terms are divisible by $a$ except for the $b^4+1$. As $\gcd(a, b) = 1$, the result similarly follows. $\blacksquare$ Now fix $r = \lfloor \sqrt a \rfloor$ and $k = a-b$. As $k \leq 2r$, we have the inequality $$\frac{k^4+1}{ab} \leq \frac{(2r)^4+1}{r^2(r^2+k)} < 16.$$Set $s \in \mathbb N$ to be the quotient. Claim. [Second Main Step] $s = 1$. Proof. Only $s = 1, 2, 5, 10, 13$ can work by Fermat Christmas. If $s$ is even, then $k$ is odd, and $4 \mid sab$, which is impossible. $s=5$ and $s=13$ don't work because $x^4 \equiv -1 \pmod s$ does not have solutions for $s \in \{5, 13\}$. $\blacksquare$ To finish, note that we have $$a+b = \sqrt{(a-b)^2+4ab} = \sqrt{4k^4+k^2+4}.$$For $4k^4+k^2+4=m^2$ to be a perfect square, we must have $\Delta = 16m^2 - 63$ be a perfect square; the only possibilities are $m = 2, 3, 8$. Of these, only $m = 2, 3$ yield valid solutions, which give the two pairs indicated above.

We claim our only solution are the edge cases $\boxed{(1,1),(1,2),(2,1)}$. Otherwise, define $n = \lfloor\sqrt{a}\rfloor=\lfloor \sqrt{b}\rfloor > 1$. Then $\gcd(a,b)=1$, and combining this with \[0 \equiv b^4+1 \equiv (a+(b-a))^4+1 \equiv (a-b)^4+1 \pmod{a},\] we find analogously that $k := \frac{(a-b)^4+1}{ab} \in \mathbb{N}$. First notice that $v_2$ tells us $k$ is odd, so we claim that we must have $k=1$. Considering a prime $p \mid k$, we get \[p \mid (a-b)^4+1 \implies \operatorname{ord}_p(a-b) = 8 \implies p \equiv 1 \pmod 8 \implies p \ge 17,\] but we also have the size bound $k < \frac{(2n)^4+1}{n^2(n^2+1)} < 16$, contradiction. We further derive a contradiction for $k=1$, as \[n^4+1 < n^2(n^2+1) \leq ab \leq (n^2+2n-1)(n^2+2n) < (n+1)^4+1. \quad \blacksquare\]

willwin4sure wrote: Find all pairs of positive integers $(a,b)$ satisfying the following conditions: $a$ divides $b^4+1$, $b$ divides $a^4+1$, $\lfloor\sqrt{a}\rfloor=\lfloor \sqrt{b}\rfloor$. Yang Liu Let $a=n^2+x$ and $b=n^2+y$ then we have $n^2+x|(x-y)^4+1$ and $n^2+y|(x-y)^4+1$. Let $d=(n^2-x,n^2-y)$ then $d|x-y$ and so $d|1$ so $d=1$. So we get that $(n^2+x)(n^2+y)|(x-y)^4+1$ or $k(n^2+x)(n^2+y)=(x-y)^4+1$ We have that $x-y=<2n$ and so $k=<16$ Also the equation $c^4=-1(mod k)$ shoyld have sollution.From here we have that $3,4,5,7,11,13$ can not devise $k$ so $k=1,2$. If $k=2$ we have that $x-y=1(mod2)$ so $(x,y)=(0,1)or(0,1)(mod(2,2))$ and here we have that $4|LHS$ contradiction. So $k=1$ an d we have that: $(n^2+x)(n^2+y)=(x-y)^4+1$ For $n=1$ we have $0<=x,y<=2$ and we get esily $(a,b)=(1,1),(2,1),(1,2)$ For $n>1$ we have that $x-y>=n$ but also $x-y<n+1$ otherwise $LHS<RHS$ so $x-y=n$ and we have: $n^2(x+y)+xy=1$ since $n>1$ we have $x=y=0$ but then no sollution. Therefore, the only solutions are $\boxed{(a,b)=(1,1),(2,1),(1,2)}$

Here is a sketch. The answers $(a, b) = (1, 1)$, $(2, 1)$, and $(1, 2)$. We can see the that the problem reduce to. $k := \frac{(a-b)^4+1}{ab} \in \mathbb{N}$, we show that $k=1$ which will give us the above answers.

https://drive.google.com/file/d/1CyRU1CsAiiRYRdbAMhO0uLmgxILiuEx7/view?usp=drivesdk it's my soln