Find all primes 'p'

Level 2

How many prime numbers 'p' are there such that p a p^{a} + p b p^{b} is a square number.


The answer is 2.

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1 solution

Jordan Cahn
Feb 8, 2019

Let p p be a prime such that p a + p b = m 2 p^a+p^b=m^2 for some non-negative integers a , b , m a,b,m . If a = b a=b then m 2 = 2 p a m^2 = 2p^a and p p must be even, so p = 2 p=2 . Set this case aside. Now, without loss of generality, let a > b a>b . Then m 2 = p a + p b = p b ( p a b + 1 ) m^2=p^a+p^b = p^b(p^{a-b}+1) . Since a b > 0 a-b>0 , p a b + 1 p^{a-b}+1 cannot be divisible by p p . But m 2 m^2 is divisible by p p and, therefore, p 2 p^2 . So b b is even. Say b = 2 c b=2c .

Note that we can now write m 2 = ( p c ) 2 ( p a b + 1 ) m^2 = (p^c)^2(p^{a-b}+1) . This implies that p a b + 1 p^{a-b}+1 is itself a perfect square. To simplify things, let's say a b = n a-b=n and p a b + 1 = k 2 p^{a-b}+1 = k^2 .

Using our new notation, we have p n + 1 = k 2 p^n+1=k^2 . Rearranging yields p n = k 2 1 = ( k 1 ) ( k + 1 ) p^n=k^2-1=(k-1)(k+1) . Since k > 0 k>0 we know that k + 1 > 1 k+1>1 . We now have two cases:

  • k 1 = 1 k-1 = 1 . In this case, k = 2 k=2 .
  • k 1 > 1 k-1>1 . Then k 1 k-1 and k + 1 k+1 must both be powers of p p , and therefore divisible by p p . But then ( k + 1 ) ( k 1 ) = 2 (k+1)-(k-1) = 2 would be divisible by p p . So p = 2 p=2 .

So either p = 2 p=2 or k = 2 k=2 . In the latter case, p n + 1 = 4 p^n+1=4 and p n = 3 p^n=3 . So p = 3 p=3 .


We have now determined that the only possible solutions are p = 2 p=2 and p = 3 p=3 . We will verify that these are indeed solutions.

  • The solution p = 3 p=3 arose in the case where 3 a b + 1 = 4 3^{a-b}+1=4 Therefore a b = 1 a-b=1 . Any such a a and b b (with b = 2 c b=2c ) will suffice, since 3 2 c + 1 + 3 2 c = ( 3 c ) 2 ( 3 + 1 ) = ( 2 3 c ) 2 3^{2c+1}+3^{2c} = (3^c)^2(3+1) = (2\cdot 3^c)^2 . One such example would be 3 1 + 3 0 = 3 + 1 = 4 = 2 2 3^1+3^0 = 3 + 1 = 4 = 2^2
  • The solution p = 2 p=2 arose when a = b a=b . This yields a solutions whenever a a is odd (say a = 2 d + 1 a=2d+1 ), since then 2 a + 2 a = 2 2 2 d + 1 = ( 4 2 d ) 2 2^a+2^a = 2\cdot2^{2d+1} = (4\cdot 2^d)^2 . One such example would be 2 1 + 2 1 = 2 + 2 = 4 = 2 2 2^1+2^1 = 2+2 = 4 = 2^2 . These are not the only solutions: if b = 2 c b=2c is even, 2 b + 3 + 2 b = 2 2 c ( 2 3 + 1 ) = ( 3 2 c ) 2 2^{b+3} + 2^b = 2^{2c}(2^3 + 1) = (3\cdot 2^c)^2 will be a solution. For example, 2 3 + 2 0 = 8 + 1 = 9 = 3 2 2^3+2^0 = 8 + 1 = 9 = 3^2 .

Thus, there are 2 \boxed{2} possible values of p p .


If we wish, we can show that these are the only classes of solutions. We have already shown that no other p = 3 p=3 solutions can exist. For p = 2 p=2 , we know that either a = b a=b or b b must be even and 2 a b + 1 2^{a-b}+1 must be a perfect square. But we have seen that the latter case implies that both k 1 k-1 and k + 1 k+1 are powers of 2 2 . This is only possible when k = 3 k=3 . Thus 2 a b + 1 = 3 2 = 9 2^{a-b}+1 = 3^2 = 9 and a b = 3 a-b=3 . So, if p = 2 p=2 , then either a = b a=b or a = b + 3 a=b+3 .

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