Substracting sides, we get $ y(y-1) - x(x-1) = m - n $, note that both $y(y-1) $ and $ x(x-1) $ are even, so $ 2 | m-n $. Now for any $k$, the system $ x^2 + y = k^2 + k + 1 $ and $ x + y^2 = k^2 + 3k + 1 $ has a solution $(k, k+1)$, it is simple to prove that this is the only solution through some estimates. So the answer is any number divisible by 2.

you miss the case $m-n=0$

Is that a solution ???? I have that there is no solution Am i true???? The proof i will give later

does anyone have solutions???

chaotic_iak wrote: For some positive integers $m,n$, the system $x+y^2 = m$ and $x^2+y = n$ has exactly one integral solution $(x,y)$. Determine all possible values of $m-n$. Dear all of You: Sorry, my prior "solution" contains a fatal error. I thus had to withdraw it - I don't want to spam our forum Good night, auj

I think I have a solution. Subtracting the two equations, we get $(x-y)(1-x-y)=m-n$. If $m-n=0$, $(x-y)(1-x-y)=m-n=0$ has infinitely many solution $(x-y)$, contradicting the uniqueness of $(x,y)$. Hence, $m \ne n$. Let $d_1=x-y$ and $d_2=1-x-y$ denote two divisors of $m-n$ such that $d_1d_2=m-n$. Solving for $y$, we get $y=\frac{1-d_1-d_2}{2}$, implying $d_1+d_2$ is odd. Thus, $m-n$ must be even, since exactly one of $d_1$ and $d_2$ must be even. But since the system has exactly one integral solution pair $(x,y)$, there exists only one solution pair to $(d_1, d_2)$, namely $(m-n,1)$. From this, $|m-n|$ must be a power of 2, otherwise there will be more than one solution to the system. Therefore, $m-n=\pm2^k$, where $k \in \mathbb{Z}^+$.