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We have an equation: $$m^2 = n^2 + m + n + 2018.$$ Find all integer pairs $(m,n)$ satisfying this equation.

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closed as off-topic by Théophile, John Omielan, Eevee Trainer, YiFan, zz20s Mar 9 at 0:39

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    $\begingroup$ Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers. $\endgroup$ – fleablood Mar 8 at 16:38
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Hint $$ (m+n)(m-n)= (m+n)+2018$$

so $$ (m+n)(m-n-1)= 2018$$

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Guide: Write $m=n+k$ for some integer $k$, then $$n^2+2nk+k^2= n^2+2n+k+2018$$

so $$ n={-k^2+k+2018\over 2(k-1)}=-{k\over 2}+{1009\over k-1}$$

If $k$ is odd then there is no solution, so $k= 2s$ so $$2s-1\mid 1009$$

Can you finish?

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Simpler start: separating variables to either side gives: $$m^2-m=n^2+n+2018$$ which then factors roughly for the variables as: $$m(m-1)=n(n+1)+2018$$

which since both pairs(m,m-1) and (n,n+1) are consecutive integers, you can divide both sides by two giving:

$$\frac{m(m-1)}{2}=\frac{n(n+1)}{2}+1009$$

But, $\frac{y(y+1)}{2}$ is the form of the y-th triangular number, so the solutions are such that 1009 is the difference of two triangular numbers $T_{\vert m-1 \vert}$ and $T_{\vert n \vert}$ . Solve for n, and m-1 .

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