# 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?

• There's the second factor. What happens if that is $0$? Nov 13 '20 at 14:07
• What do u mean by "There's the second factor" Nov 13 '20 at 14:12
• You wrote "factorizing". I suppose you got more than one factor doing that. Nov 13 '20 at 14:13
• Oh thanks that second factor gives m = n and solves the question ;) Nov 13 '20 at 14:17
• How can i end my question ? Nov 13 '20 at 14:18

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

• Not clear precisely which way you deduced $\,m=n,\,$ but if it is from $\,m\neq n\Rightarrow\, -2(m+n) = 1\,$ then this proof by parity (divisibility) contradiction generalizes widely - see my answer where I shows how to extend it from the divisor $2$ to $d.\ \$ Nov 14 '20 at 2:14
• @Bill Dubuque product of two integers is 0 implies that atleast one of them is 0. Since $m$ and $n$ are naturals, $2m+2n+1$ cannot be $0$. Dec 12 '20 at 13:50
• Yes, that's essentially the proof by parity that I surmised you used. Dec 12 '20 at 15:17

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 $$m,n \in \Bbb Z$$.

• Nice method but u just complicated it ,if x is not equal to y then x + y = -0.5 which is not true Nov 13 '20 at 14:26
• @AdhirajSinghBrar You're right. When I first wrote it I was trying to figure out why $m-n$ would be a perfect square... Nov 13 '20 at 14:27
• @AdhirajSinghBrar I made the proof more succinct now Nov 13 '20 at 14:31
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.
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.