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Saudi Arabian Mathematical Competitions

Number theory Difficulty 5.6 AIME, harder Prove it Saudi Arabia

Find all positive integers nn for which 1−5n+52n+11-5^{n}+5^{2 n+1} is a perfect square.

Solution

Assume that 1−5n+52n+1=m21-5^{n}+5^{2 n+1}=m^{2}, for a positive integer mm. We have
5n(5n+1−1)=(m−1)(m+1). 5^{n}\left(5^{n+1}-1\right)=(m-1)(m+1) .
Because (m+1)−(m−1)=2(m+1)-(m-1)=2, the number 55 cannot divide both m−1m-1 and m+1m+1. Therefore, we have two cases:

First case when 55 divides m−1m-1. In this case, there exists a positive integer kk such that m−1=5nkm-1=5^{n} k and 5n+1−1=(m+1)k=(5nk+2)k5^{n+1}-1=(m+1) k=\left(5^{n} k+2\right) k. We, therefore, obtain the equation
5n(5−k2)=2k+1 5^{n}\left(5-k^{2}\right)=2 k+1
For k=1k=1, this equation has no solution. For k=2k=2, n=1n=1 is a solution for the equation. For k≥3k \geq 3 this equation has no solution since its left hand side is negative.

Second case when 55 divides m+1m+1. In this case, there exists a positive integer kk such that m+1=5nkm+1=5^{n} k and 5n+1−1=(m−1)k=(5nk−2)k5^{n+1}-1=(m-1) k=\left(5^{n} k-2\right) k. We, therefore, obtain the equation
5n(k2−5)=2k−1 5^{n}\left(k^{2}-5\right)=2 k-1
This equation has no solution since for k=1,2k=1,2 or 33 the left hand side is either negative or even, and for k≥4k \geq 4 we have
5n(k2−5)>5(k+2)(k−3)≥5k+10>2k−1. 5^{n}\left(k^{2}-5\right)>5(k+2)(k-3) \geq 5 k+10>2 k-1 .
Therefore, the unique solution to this problem is n=1n=1 since we have
1−5+53=112. 1-5+5^{3}=11^{2} .

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