Find the number of primes \(p\) for which there exists a positive integer \(m\) that is of the form \(12k+2\) such that \(48m(47m+1)\) is \(p\) times a perfect square.
Solution
Answer: \(0\)
Suppose \(48m(47m+1)=p x^2\). Since \(m=12k+2\), any common divisor of \(48m\) and \(47m+1\) divides \(48\), while \(47m+1\) is divisible by neither \(2\) nor \(3\). Thus the two factors are coprime.
One factor must be a square and the other \(p\) times a square. But \(m\equiv2\pmod4\), so \(v_2(48m)=5\); consequently \(48m\) cannot be a square. We must have \(47m+1=b^2\).
Reducing modulo \(12\) gives \(b^2\equiv-m+1\equiv11\pmod{12}\), whereas square residues modulo \(12\) are \(0,1,4,9\). This contradiction shows that no primes work.