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

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

First, we observe that

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

m+n+6=6\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)(m,n) that satisfies the equation,


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

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

m272mn+1290=0m^2-72m-n+1290=0 and it's a quadratic equation in mm which has real solutions if the discriminant, Δ0\Delta \ge 0.

Δ=7224(1)(n+1290)0\Delta = 72^2-4(1)(-n+1290) \ge 0

24+4n0\Rightarrow 24+4n \ge 0

n6\Rightarrow n \ge -6 which is trivial since n>0n>0.

Also, using the quadratic formula, we get,

m=72±24+4n2=36±6+nm=\frac{72\pm\sqrt{24+4n}}{2}=36\pm\sqrt{6+n} since m<36m<36 we shall take the negative sign, this leaves

m=366+nm=36-\sqrt{6+n} . Thus, let 6+n=k<36\sqrt{6+n}=k <36, where kk is a positive integer, then n=k26n=k^2-6.

For n,m>0n,m>0, k=3,4,5,...,35k=3,4,5,...,35, at last, there are 3333 pairs of positive integers (m,n)(m,n).

Note by ChengYiin Ong
1 year, 8 months ago

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Great. Thanks 😊

A Former Brilliant Member - 1 year, 8 months ago

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