Let $m, n ∈ \mathbb{N}$ such that $2m^{2} + m = 2n^{2} + n$ , then prove that $m-n$ is a perfect square . After simply factorizing the equation given in the question I got $\mathbf{m + n = -0.5}$
. But the question mentioned $m, n ∈ \mathbb{N}$ . Then how $m + n = -0.5$ ?
Did I do some mistake or is the question wrong?
 A: Yes. $m-n$ is a perfect square.
Just observe that your condition implies that
$$ (m-n)(2m+2n+1)=0$$
Since $m$ and $n$ are natural numbers, we must have $m-n = 0$ which is a perfect square
A: As you conclude in the comments, the point is that we must have $m = n$.
Here's an alternative proof: suppose for contradiction that $m \neq n$; in particular take $m > n$. Let $k = 2m^2 + m$. We see that $m$ and $n$ are two solutions to the quadratic equation
$$
2x^2 + x - k = 0.
$$
The sum of the roots of $ax^2 + bx + c = 0$ is given by $-\frac{b}{a} = - \frac 12$, which is to say that $m + n$ is not an integer. This is impossible since $m,n \in \Bbb Z$.
A: An alternative way to approach the same result: For nonnegative values of $x$, the function $f(x)=2x^2+x$ is strictly increasing, so the only way to have $f(m)=f(n)$ is if $m=n$.  And then of course $m-n=0$ is a perfect square.
A: Special case of that below, namely $\, d=2,\ f = 2x^2\!+\!x$.
Theorem $\ $ If $\,f(x)\in\Bbb Z[x]\,$ and $\!\bmod d\!:\ f\equiv bx\!+\!c,\, \color{#c00}{b\not\equiv 0}\,$ then $\,f(m)=f(n)\Rightarrow m = n$
Proof $\,\ 0 = f(m)\!-\!f(n) = (m\!-\!n)\:\!k,\,$ for $\,k = \dfrac{f(m)-f(n)}{m-n}\in\Bbb Z,\,$ thus if $\,m\neq n\,$ then $\,k = 0\,$ thus $\bmod d\!:\,\ \color{#c00}0 = k \equiv \dfrac{bm\!+\!c-(bn\!+\!c)}{m-n}\equiv \color{#c00}b,\,$ contra hypothesis.
