# $\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+...}}}}}}=6$

Let's solve a fun problem relating infinite square roots,

Find the number of pairs of positive integers (m, n) such that $\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+...}}}}}}=6$ .

First, we observe that

$\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+...}}}}}}=6$ is equal to

$\sqrt{m+\sqrt{n+6}}=6$ since the expression is infinite, and we can ONLY DO THIS IFF IT'S INFINITE, take note. So, the problem reduces to find the number of pairs of positive integers $(m,n)$ that satisfies the equation,

$\sqrt{m+\sqrt{n+6}}=6$

We can first say that $m<36$ since $\sqrt{n+6}$ and $n$ are positive.

Let's square both sides and simplify the expression a little bit, and at last it will become

$m^2-72m-n+1290=0$ and it's a quadratic equation in $m$ which has real solutions if the discriminant, $\Delta \ge 0$.

$\Delta = 72^2-4(1)(-n+1290) \ge 0$

$\Rightarrow 24+4n \ge 0$

$\Rightarrow n \ge -6$ which is trivial since $n>0$.

Also, using the quadratic formula, we get,

$m=\frac{72\pm\sqrt{24+4n}}{2}=36\pm\sqrt{6+n}$ since $m<36$ we shall take the negative sign, this leaves

$m=36-\sqrt{6+n}$ . Thus, let $\sqrt{6+n}=k <36$, where $k$ is a positive integer, then $n=k^2-6$.

For $n,m>0$, $k=3,4,5,...,35$, at last, there are $33$ pairs of positive integers $(m,n)$.

Note by ChengYiin Ong
1 week, 2 days ago

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