Croatia_2018

We are given the equation $$n^2-6n=m^2+m-10.$$ Let's rewrite it as $$n^2-6n-m^2-m+10=0.$$ Group terms: $$n^2-m^2-6n-m+10=0.$$ Recall that $n^2-m^2=(n-m)(n+m)$, so $$(n-m)(n+m)-6n-m+10=0.$$ Let us solve for $n$ in terms of $m$.

Alternatively, move all terms to one side: $$n^2-6n-m^2-m+10=0$$ $$n^2-6n=m^2+m-10$$ This is a quadratic in $n$: $$n^2-6n-(m^2+m-10)=0$$ The discriminant must be a perfect square for integer solutions: $$\Delta =36+4(m^2+m-10)=36+4m^2+4m-40=4m^2+4m-4=4(m^2+m-1)$$ So $4(m^2+m-1)$ must be a perfect square, say $k^2$ for some integer $k$: $$4(m^2+m-1)=k^2$$ So $$k^2-4m^2-4m+4=0$$ $$k^2-4m^2-4m+4=0$$ $$k^2=4m^2+4m-4$$ $$k^2=4(m^2+m-1)$$ So $m^2+m-1$ must be a perfect square times $1/4$. But $k$ must be even, so let $k=2t$ for some integer $t$: $$(2t{)}^2=4(m^2+m-1)$$ $$4t^2=4(m^2+m-1)$$ $$t^2=m^2+m-1$$ So $m^2+m-1=t^2$ for some integer $t$.

Now, rearrange: $$m^2+m-1-t^2=0$$ $$m^2+m-t^2=1$$ This is a quadratic in $m$: $$m^2+m-(t^2+1)=0$$ The discriminant must be a perfect square: $$\Delta =1^2-4\cdot 1\cdot (-t^2-1)=1+4t^2+4=4t^2+5$$ So $4t^2+5$ must be a perfect square, say $s^2$: $$s^2=4t^2+5$$ So $s^2-4t^2=5$

This is a Pell-type equation.

Let us solve $s^2-4t^2=5$ in integers.

Try small integer values for $t$: - $t=0$: $s^2=5$ (no integer solution) - $t=1$: $s^2=4+5=9$, $s=3$ - $t=-1$: $s^2=4+5=9$, $s=3$ - $t=2$: $s^2=16+5=21$ (no integer solution) - $t=-2$: $s^2=16+5=21$ (no integer solution) - $t=3$: $s^2=36+5=41$ (no integer solution) - $t=-3$: $s^2=36+5=41$ (no integer solution) - $t=5$: $s^2=100+5=105$ (no integer solution)

So only $t=1$ and $t=-1$ work, with $s=3$.

Now, recall $m^2+m-t^2=1$ For $t=1$: $$m^2+m-1=1$$ $$m^2+m=2$$ $$m^2+m-2=0$$ $$m=\frac{-1\pm \sqrt{1+8}}{2}=\frac{-1\pm 3}{2}$$ So $m=1$ or $m=-2$

For $t=-1$: Same as above, $m=1$ or $m=-2$

Now, recall $n$ is given by the quadratic: $$n^2-6n=m^2+m-10$$ For $m=1$: $$n^2-6n=1+1-10=-8$$ $$n^2-6n+8=0$$ $$n=\frac{6\pm \sqrt{36-32}}{2}=\frac{6\pm 2}{2}=4,2$$ For $m=-2$: $$n^2-6n=4-2-10=-8$$ So same as above, $n=4,2$

Therefore, the integer solutions are: $$(m,n)=(1,2),(1,4),(-2,2),(-2,4)$$

Thus, all integer pairs $(m,n)$ such that $n^2-6n=m^2+m-10$ are: $$(1,2),(1,4),(-2,2),(-2,4)$$