Let $k= \frac{(m+n)^2}{4(m(m-n)^2 + 1)}$. If $m \not = n$, then $m(m-n)^2 + 1 \not = 1$, so this number has a prime divisor $p$. But since $k$ is assumed to be an integer we have that $p \vert m(m-n)^2 + 1 \implies p \vert (m+n)^2$. Multiplying the first relation by $(m+n)^2$ and subtracting the second one from it we get: $$p \vert m(m-n)^2(m+n)^2 + (m+n)^2 - (m+n)^2 \implies p \vert m(m-n)^4$$ But this is impossible, as $p$ doesn't show up into the prime representation of neither $m$, nor $(m-n)$, since otherwise initial relation implies $p \vert 1$, which is impossible. So we get a contradiction and therefore $k \in \mathbb{Z} \implies m=n$ Therefore $k=\frac{(m+m)^2}{4m(m-m)^2 + 4} = \frac{(2m)^2}{4} = m^2$. Q.E.D.

Sorry how can you conclude that $m(m-n)^2 + 1 \not = 1$ has a prime divisor??

Miracle_ wrote: Sorry how can you conclude that $m(m-n)^2 + 1 \not = 1$ has a prime divisor?? $m(m-n)^2 + 1$ is obviously an positive integer and it's a well-known fact that any positive integer other than $1$ is either a prime number or a product of finitely many prime numbers. As I said it's well-known fact, but if you want a proof, you can easily do that by strong induction.

Stefan4024 wrote: Let $k= \frac{(m+n)^2}{4(m(m-n)^2 + 1)}$. If $m \not = n$, then $m(m-n)^2 + 1 \not = 1$, so this number has a prime divisor $p$. But since $k$ is assumed to be an integer we have that $p \vert m(m-n)^2 + 1 \implies p \vert (m+n)^2$. Multiplying the first relation by $(m+n)^2$ and subtracting the second one from it we get: $$p \vert m(m-n)^2(m+n)^2 + (m+n)^2 - (m+n)^2 \implies p \vert m(m-n)^4$$ But this is impossible, as $p$ doesn't show up into the prime representation of neither $m$, nor $(m-n)$, since otherwise initial relation implies $p \vert 1$, which is impossible. So we get a contradiction and therefore $k \in \mathbb{Z} \implies m=n$ Therefore $k=\frac{(m+m)^2}{4m(m-m)^2 + 4} = \frac{(2m)^2}{4} = m^2$. Q.E.D. $5|4+1$ $5|1(4-1)^2+1$ $m=1$ ,$n=4$

Stefan4024 wrote: $$p \vert m(m-n)^2(m+n)^2 + (m+n)^2 - (m+n)^2 \implies p \vert m(m-n)^4$$ You mistake $(m-n)^2(m+n)^2$ for $(m-n)^4$

Official Solution: If $m=n$ then $k=m^2$ is a perfect square. Let $m\neq n$. Put $x=m+n$ and $y=n-m$ for $n>m$, and $y=m-n$ for $n<m$. In both cases, $x\equiv y \pmod2$. Since $m=\dfrac{x-y}{2}$ for $n>m$ and $m=\dfrac{x+y}{2}$ for $n<m$, we have $k=\dfrac{x^2}{2(x\pm y)y^2+4}$. Thus, we have a second degree equation$$x^2-2ky^2x\pm 2ky^3-4k=0$$Since $x$ is an integer, the discriminant of this equation should be a perfect square:$\dfrac{\Delta}{4}=k^2y^4\mp2ky^2+4k$ is a perfect square. If $y=1$ then $k=\dfrac{x^2}{2x+2}$ or $k=\dfrac{x^2}{2x+6}$. In the first case, $2k=x-1+\dfrac{1}{x+1}$ is not an integer. In the second case, $2k=x-3+\dfrac{9}{x+3}$ and the only possibility is $x=6$, but in this case $x\equiv y\pmod2$ is not held. Let $y\neq1$. Let us prove the following two inequalities$$(y^2k\mp y-1)^2<\dfrac{\Delta}{4}<(y^2k\mp y+1)^2$$ The proof of the first inequality: Note that $(y^2k\mp y-1)^2=k^2y^4+y^2+1\mp 2ky^3-2ky^2\pm2y<k^2y^4\mp2ky^3+4k$. Equivalently, $(y\mp 1)^2<k(2y^2+4)$, which is equivalent to $k>\dfrac{(y\mp1)^2}{2y^2+4}$. Since $2y^2+4>(y\pm1)^2\Longleftrightarrow(y\mp1)^2+2>0$ and we are done. The proof of the second inequality: Note that $k^2y^4\mp2ky^3+4k<(y^2k\mp y+1)^2=k^2y^4+y^2+1\mp2ky^3+2ky^2\mp 2y$. Equivalently, $4k<(y\mp1)^2+2ky^2$ which holds for all $y\ge2$. Now since $y\neq1$, we get $\dfrac{\Delta}{4}=(y^2k\mp y)^2$. Thus, $4k=y^2$ and $k$ is a perfect square. Note: It can be shown that $k$ is an integer number for infinitely many pairs $(m,n)$.

My Solution: if you multiple both sides by $4m(m-n)^2+4$ and then rewrite everything ordered by its $n$ degree and factor it like a second degree equation of $n$ its and compute $\Delta=b^2-4ac$ you will get $$\Delta=16k(4m^3-4m+1)$$now we will show that $gcd(k,4m^3-4m+1)=1$ if $p|k$ then $p|m+n$ now we have $$4m(m-n)^2+4 \equiv 4m(2m)^2+4 \equiv 16m^3+4 \pmod p$$but we had $4m^3 \equiv 4m-1 \pmod p $ hence $16m^3+4 \equiv 16m \pmod p$ this shows that if p divides $4m(m-n)^2+4$ then $p|m$ which is clearly impossible and so $gcd(k,4m^3-4m+1)=1$ and so k is a square.

Eray wrote: Official Solution: $\dfrac{\Delta}{4}=k^2y^4\mp2ky^2+4k$ is a perfect square. Of course, it is wrong: $\dfrac{\Delta}{4}=k^2y^4\mp2ky^3+4k$. Discriminant in solution of matinyousefi is also wrong, it really equals $16k(4m^3-4mk+1)$. So we still have no solution

Pavel, how about this? Note that, $k>0$. First, start by checking that, if $m=n$; then, $k=(m+n)^2/4$, in particular, it is a perfect square. Also, note that, $4\mid (m+n)^2$; and thus, $m+n\equiv m-n\equiv 0 \pmod{2}$. Now, we let, $$ a=\frac{m+n}{2}\quad\text{and}\quad b=\frac{m-n}{2}. $$With this, the requirement that $k$ is an integer, reads as, $$ k=\frac{a^2}{4b^2(a+b)+1}\implies a^2-(4kb^2)a-(4kb^3+b)=0. $$We interpret this as a quadratic in $a$. If $a\in\mathbb{Z}$; then we must have $\Delta$ (discriminant) to be a perfect square, and therefore, $$ \Delta = 16k^2b^4 +16kb^3 +4k = 4(4k^2b^4+4kb^3+k). $$In particular, we focus on $4k^2b^4+4kb^3+k$; and investigate when it is a perfect square. Suppose that $b>0$. We have, $$ (2kb^2+b-1)^2<4k^2b^4+4kb^3+k<(2kb^2+b+1)^2, $$which can be checked manually, and easily. Hence, in this case, $$ 4k^2b^4+4kb^3+k = (2kb^2+b)^2 = 4k^2b^4+4kb^3+b^2 \implies k=b^2, $$as desired. Now, suppose that $b<0$. We let $c=-b$; and get $c>0$, and rewrite everything in terms of $c$, as, $$ (2kc^2-c-1)^2 <4k^2c^4-4kc^3+k <(2kc^2-c+1)^2, $$and therefore, $$ 4k^2c^4-4kc^3+k=(2kc^2-c)^2 =4k^2c^4-4kc^3+c^2 \implies k =c^2. $$In particular, $k=(m-n)^2/4$; and we are done.

Yes, your solution is similar to official one. Now I realize that square bounding works here.