<s>[INST] Basic Number Theory Question involving quadratic equations and squares Question : Let  $m,n\in \mathbb{N}$  such that $2m^2 + m = 2n^2 + n$, then prove that $m - n$ and $2m + 2n + 1$
are perfect squares.
$\begin{align}2m^2-2n^2 &=n-m\\
-2(n-m)(m+n) &= n-m \\\end{align}$
If $m-n \neq 0
$\begin{align}\Rightarrow 2(m+n) &=-1  \\
\therefore 2m+2n+1 &=-1+1=0=0^2 \text{ } \blacksquare\\
\end{align}
If $m-n=0$, then $m-n=0=0^2 \text{ } \blacksquare$
Also, Let $2m^2 + m = 2n^2 + n=k$
Clearly, $m,n$ are solutions to the quadratic equation $2x^2+x-k=0$
This implies that $mn=\frac{-k}{2}$
,Implying either $m$ or $n$ is negative which is incorrect as we are given that $m,n \in N$
Therefore to avoid this contradiction it must be that $m=n$ and there exists another root to the equation $\alpha \in \mathbb{Q}$.
Thus $m-n=0=0^2 \text{  } \blacksquare $
Please just check my solution and tell me if it is correct, if not suggest changes please i wish to learn. Also my main doubt is in the first half of the proof i know it does look very sketchy that the sum of two natural numbers is negative. Thanks a lot for your help and understanding <$3$.
 [/INST] $$ 2m^2 + m = 2 n^2 + n $$
$$ 2m^2 - 2n^2 + m - n = 0 $$
$$ 2(m^2-n^2) +(m-n) = 0  $$
$$ 2(m+n)(m-n) + (m-n) = 0 $$
$$ ( 2m+2n )(m-n) + 1 (m-n) = 0 $$
$$ (2m+2n+1)(m-n) = 0  $$
With integers, the left factor is odd so nonzero, thus $m=n$
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