February 15, 2023
Problem 4803 Crux Mathematicorum Vol. 49, No. 1
Find all non-negative integers \(a\), \(b\), \(c\) and pairs \((p, q)\) of prime numbers satisfying \[ p^{2a} + q^{2b} = (2c+1) ^ 2.\]
(Proposed by Nguyen Viet Hung)
Since \((2c+1) ^ 2\) is an odd number, one and only one of \((p, q)\) must be \(2\) otherwise \(p^{2a} + q^{2b}\) will be an even number. Let \(p=2\) and \(q\) be the prime greater than \(2\), we have \[ q^{2b} = (2c+1)^2 - 2^{2a} = (2c + 1 + 2^a)(2c + 1 - 2^a). \]
Let \(s = 2c + 1 + 2^a\) and \(t = 2c + 1 - 2^a\), we have \(s > t > 0\). Since \(s - t = 2^{a+1}\) which is not divisible by \(q\), it follows that \(t = 1\). (If \(t\) is not 1, then both \(s\) and \(t\) are multiples of \(q\), which implies that \(s-t\) is also multiple of \(q\).) Therefore \(2^a = 2c\), and it follows that \[q^{2b} = (2c+1)^2 - (2c)^{2} = 4c + 1 = 2^{a+1} + 1.\] We claim that \(q\) must be \(3\). If \(q\) is a prime greater than \(3\), then \(q\) is either \(3k+1\) or \(3k-1\) for some integer \(k\), which implies that \(q^{2b}\equiv 1 \pmod 3\). However \(2^{a+1} + 1 \equiv {(-1)}^{a+1} + 1 \pmod 3\), but \({(-1)}^{a+1} + 1\) is either \(0\) (when \(a\) is even) or \(2\) (when \(a\) is odd). Thus \(q\) must be \(3\).
Then we have \(3^{2b} = 2^{a+1} + 1\) or \((3^b + 1) (3^b -1 ) = 2^{a+1}\), i.e., both \(3^b + 1\) and \(3^b - 1\) are power of \(2\), let \(3^b+1 = 2^m\), \(3^b-1 = 2^n\), where \(m>n\), the only pair of \((m, n)\) such that \(2^m - 2^n = 2\) is \((m=2, n=1)\).
Therefore \(\boxed{a=2, b=1, c=2, p=2, q=3}\) or \(\boxed{a=1, b=2, c=2, p=3, q=2}\) are the only solutions satisfying \[ p^{2a} + q^{2b} = (2c+1) ^ 2\] where \(a\), \(b\), \(c\) are non-negative integers and \((p, q)\) are prime numbers.