How many prime numbers 'p' are there such that + is a square number.
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Let p be a prime such that p a + p b = m 2 for some non-negative integers a , b , m . If a = b then m 2 = 2 p a and p must be even, so p = 2 . Set this case aside. Now, without loss of generality, let a > b . Then m 2 = p a + p b = p b ( p a − b + 1 ) . Since a − b > 0 , p a − b + 1 cannot be divisible by p . But m 2 is divisible by p and, therefore, p 2 . So b is even. Say b = 2 c .
Note that we can now write m 2 = ( p c ) 2 ( p a − b + 1 ) . This implies that p a − b + 1 is itself a perfect square. To simplify things, let's say a − b = n and p a − b + 1 = k 2 .
Using our new notation, we have p n + 1 = k 2 . Rearranging yields p n = k 2 − 1 = ( k − 1 ) ( k + 1 ) . Since k > 0 we know that k + 1 > 1 . We now have two cases:
So either p = 2 or k = 2 . In the latter case, p n + 1 = 4 and p n = 3 . So p = 3 .
We have now determined that the only possible solutions are p = 2 and p = 3 . We will verify that these are indeed solutions.
Thus, there are 2 possible values of p .
If we wish, we can show that these are the only classes of solutions. We have already shown that no other p = 3 solutions can exist. For p = 2 , we know that either a = b or b must be even and 2 a − b + 1 must be a perfect square. But we have seen that the latter case implies that both k − 1 and k + 1 are powers of 2 . This is only possible when k = 3 . Thus 2 a − b + 1 = 3 2 = 9 and a − b = 3 . So, if p = 2 , then either a = b or a = b + 3 .