Peter wrote: Let $n$ be a positive integer such that $2+2\sqrt{28n^{2}+1}$ is an integer. Show that $2+2\sqrt{28n^{2}+1}$ is the square of an integer. Let's start with this one. I solved it last year in ML . We have that $28n^{2}+1=m^{2}$,for an $m\in{N}$ <=> $m^{2}-28n^{2}=1$ this is a particular case of Pell's equation, and I found out that the smallest pair of solutions is $(m_{0};n_{0})=(127: 24)$=> $(m;n)=(\frac{(127+48\sqrt{7})^{k}+(127-48\sqrt{7})^{k}}{2};\frac{(127+48\sqrt{7})^{k}-(127-48\sqrt{7})^{k}}{2})$ for all $k=0;1;2;...$; Now \[2+2\sqrt{28n^{2}+1}=2m+2=(127+48\sqrt{7})^{k}+(127-48\sqrt{7})^{k}+2=((8+3\sqrt{7})^{k}+(8-\sqrt{7})^{k})^{2}\]=> $2+2\sqrt{28n^{2}+1}=((8+3\sqrt{7})^{k}+(8-\sqrt{7})^{k})^{2}$and that is easy to see that $((8+3\sqrt{7})^{k}+(8-\sqrt{7})^{k})\in{N}$ so we are done!

Peter wrote: Let $n$ be a positive integer such that $2+2\sqrt{28n^{2}+1}$ is an integer. Show that $2+2\sqrt{28n^{2}+1}$ is the square of an integer. One may slightly generalize the above problem: [Proposition] Let $p \equiv-1(mod \; 4)$ be a prime. Suppose that $2+2\sqrt{4p n^{2}+1}$ is an integer for some integer $n$. Then, $2+2\sqrt{4p n^{2}+1}$ is the square of an integer. [Proof] Since $4p n^{2}+1$ is an odd integer, we write \[4p n^{2}+1=(2u+1)^{2}\;\; \text{or}\;\; pn^{2}=u(u+1),\] where $u \in Z$. Then, we have $p \; \vert \; u$ or $p \; \vert \; u+1$ because $p$ is prime. However, $p \; \vert \; u+1$ is impossible. Assuming that $p \; \vert \; u+1$, we see that $gcd(u, u+1)=1$ and $u(u+1)=pn^{2}$ imply that $(u,u+1)=(a^{2}, pb^{2})$, $a, b \in Z$ so that \[a^{2}-pb^{2}=-1 \;\; \text{or}\;\; a^{2}\equiv-1 (mod \; p).\] This contradicts for the fact that $p \equiv-1(mod \; 4)$ or that $-1$ is a quadratic nonresidue modulo $p$. Hence, we conclude that $p \; \vert \; u$. In this case, we get $(u,u+1)=(pc^{2}, d^{2})$ for some $c, d \in Z$. It therefore follows that \[2+2\sqrt{4p n^{2}+1}=2+2(2u+1)=2+2(2d^{2}-1)=(2d)^{2}.\]

Notably, this question recently appeared in British MO Round 1, with 28 replaced with 12. Naturally, I complained that such a well known problem was used :yes:

suppose 28n^2+1=m^2, m is an integer then 7(2n)^2=(m+1)(m-1), and only one of the factor on the RHS divides 7. in fact, m-1 is a factor of 7, but not for (m+1). For the sake of contradiction, if m+1 is a factor of 7, then (2n)^2=5(mod 7), a blatant contracdition since 5 is not a quadratic residue in mod 7 Now, put m=14k+1 (n^2)=k(7k+1), and since k and 7k+1 are coprime, they must be perfect squares so 2+ 2 (28n^2+1)^0.5=4(7k+1) is a perfect square. For the case of 12 instead of 28 it works with very similar steps

We note that 2+sqrt{(28n^2+1){<?>}} is an even integer. So, 28n<?>^2+1 is a perfect sqaure of an odd integer say m.{<?>}One can easily see that 28n<?>^2=(m-1)(m+1) and 7n<?>^2=(m-1)/2*(m+1)/2 Here, m-1=14a<?>^2{<?>} and m+1=2b<?>^2{<?>} or m-1=2b<?>^2{<?>} and m+1=14a<?>^2{<?>} If the second condition is correct then we get the contradiction b<?>^2{<?>}is congruent to 7 modulo -1. Hence the first condition is correct and from that the proof follows. Sorry no Latex

Let a be a rational number such that 11+11(11a^2+1)^.5 is an odd integer, then it must be a perfect square.

A nice solution of the above problem can be found in the book "Mathematical Olympiad Challenges" by Titu Andreescu and Razvan Gelca in pages 71 and 217.

manjil wrote: Sorry no Latex Maybe it's not bad that you learn it, your solutions are really hard to read. I had to write it in LaTeX completely before I could read it. With LaTeX, manjil wrote: We note that $ 2 + 2\sqrt {28n^2 + 1}$ is an even integer. So, $ 28n^2 + 1$ is a perfect sqaure of an odd integer say $ m$. One can easily see that $ 28n^2 = (m - 1)(m + 1)$ and $ 7n^2 = \frac {m - 1}{2}\frac {m + 1}{2}$. Here,$ m - 1 = 14a^2$ and $ m + 1 = 2b^2$ or $ m - 1 = 2b^2$ and $ m + 1 = 14a^2$. If the second condition is correct then we get the contradiction $ b^2\equiv - 1\pmod{7}$. Hence the first condition is correct and from that the proof follows.

Let $28n^2+1=k^2$. Then we desire to prove that when $k$ is an integer, $2+2k$ is a perfect square also. Therefore we need $k+1=2m^2$ also. If $28n^2+1=k^2$, then we get $k^2-28n^2=1$. Therefore we need $k\equiv 1\pmod 2$ so let it be $2k'-1$. We now desire $4k'=m^2$. Substitute $k=2k'-1$ to give us $(2k'-1)^2-28n^2=1$ or $4k'^2-4k'-28n^2=0$. Therefore $k'^2-k'=7n^2$. We desire to prove that $k'$ is a perfect square from this equation. Since $\gcd(k', k'-1)=1$, we are going to have to have $k'(k'-1)=7n^2$ implying that $7|k'$ or $7|(k'-1)$. The first case gives us $k'=7k_2$ or therefore $7k_2(7k_2-1)=7n^2$ or $k_2(7k_2-1)=n^2$. We need both $k_2$ and $7k_2-1$ to be perfect squares, let them be $a^2$ and $b^2$. Therefore $7a^2-1=b^2$ or $7a^2-b^2=1$. However, this implies $b^2+1\equiv 0\pmod 7$ which is impossible by taking $b=0, 1, 2, 3, 4$. The second case gives us $k'=7k_2+1$. This gives us $(7k_2+1)(7k_2)=7n^2$ or $k_2(7k_2+1)=n^2$. Since $\gcd(k_2, 7k_2+1)=1$, we need $k_2=a^2, 7k_2+1=b^2$. Therefore $7a^2+1=b^2$ or $b^2-7a^2=1$. The trivial solution is $b=8, a=3$ and there are other solutions to this equation. Since we have $7k_2+1$ being a perfect square being the only way we can have solutions, and $7k_2+1=k'$, our proof is complete $\blacksquare$.

See "An Excursion in Mathematics"

Peter wrote: Let $n$ be a positive integer such that $2+2\sqrt{28n^2 +1}$ is an integer. Show that $2+2\sqrt{28n^2 +1}$ is the square of an integer. Similar problem Let $a\in\mathbb{Q}$. If $$11+11\sqrt{11a^2+1}$$is an odd integer, prove that it is a perfect square

my solution is pretty much like all solutions above but I still wanna post it $2+2\sqrt{28n^2 +1}\in N \implies 28n^2 +1=(2k+1)^2$ for some natural number $k$ which implies $ 28n^2 =(2k+1)^2 - 1 \implies 7n^2=k\times (k+1)$ and owing to the fact that $(k,k+1)=1 \implies $ Case 1 : there exist natural numbers $s ,t$ such as that: $k+1=7\times t^2, k=s^2$ Case 2 : there exist natural numbers $s ,t$ such as that: $k+1=t^2 , k=7 \times s^2$ suppose that the first case happens $\implies k+1 = s^2 +1 = 7 \times t^2 \implies 7 | s^2 +1 $ because $7 \equiv-1 (mod 4) \implies 7|1$ which a contradiction so the second case is true and $k+1$ is perfect square as we were asked tor prove!

If $2+2\sqrt{28a^2+1}\in \mathbb Z$ then there must exist an integer $b>0$ such that $$28a^2+1=b^2 \qquad (1)$$Notice that this is a Pell equation. We can check that the smallest solution to this is $(a,b)=(24,127)$. Now consider the sequences $127=x_0<x_1<x_2<\dots$ and $24=y_0<y_1<y_2<\dots$ such that $$ x_n+y_n\sqrt{28}=\left(127 +24 \sqrt{28} \right)\left (x_{n-1}+y_{n-1}\sqrt{28}\right) \iff \begin{cases} x_n=127x_{n-1}+672y_{n-1} \\ y_n=127y_{n-1}+24x_{n-1} \end{cases} \qquad (2)$$Then, it is known that if $(a_0,b_0)$ is a solution to $(1)$ then there is a $k$ such that $(a_0,b_0)=(y_k,x_k)$. $(2)$ gives us $x_0=127,x_1=32257$ and $$x_{n+1}=254x_n-x_{n-1}, \quad n=1,2,\dots.$$Solving this recurrence we get that $$x_n=\frac 12 \left (\left(127-48\sqrt{7}\right)^{n+1}+\left(127+48\sqrt{7}\right)^{n+1}\right)$$for all $n$. Note that \begin{align*} \sqrt{\frac{x_n+1}2} & = \frac 12 \sqrt{\left(127-48\sqrt{7}\right)^{n+1}+\left(127+48\sqrt{7}\right)^{n+1}+2} \\ & = \frac 12 \sqrt{\frac{1}{\left(127+48\sqrt{7}\right)^{n+1}}+2+\left(127+48\sqrt{7}\right)^{n+1}}\\ & = \frac 12 \left(\frac{1}{\left(127+48\sqrt{7}\right)^{(n+1)/2}}+\left(127+48\sqrt{7}\right)^{(n+1)/2}\right) \\ & = \frac 12 \left (\left(127-48\sqrt{7}\right)^{(n+1)/2}+\left(127+48\sqrt{7}\right)^{(n+1)/2}\right) \in \mathbb Z\end{align*}The last assertion is true because of the binomial theorem. In the case $n$ is even, just note that $\sqrt{127-48\sqrt 7}=8-3\sqrt 7$ and $\sqrt{127+48\sqrt 7}=8+3\sqrt 7$ then use the binomial theorem. So it follows that $x_n$ and thus $b$ must be of the form $2k^2-1$. So, $$2+2\sqrt{28a^2+1}=2+2b=2+2(2k^2-1)=(2k)^2,$$as desired.