Primes' Times!

Find the sum of all the prime numbers p p for which 2 p 1 1 p \frac{2^{p-1}-1}{p}

is a perfect square


The answer is 10.

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1 solution

For p p = 2 the 2 p 1 p \frac{2^{p-1}}{p} is not an even integer and so we can assume p p is an odd prime. Then by Fermat's Little Theorem 2 p 1 1 2^{p-1} -1 is divisible by p p . Suppose that for some integer a a , 2 p 1 1 2^{p-1}-1 = = p a 2 pa^2 .Since p p is odd we have that ( 2 ( p 1 ) / 2 1 (2^{(p-1)/2}-1 ) ( 2 ( p 1 ) / 2 + 1 ) (2^{(p-1)/2}+1) = = p a 2 pa^2 Both the factors on the L.H.S are odd and hence they are relatively prime to each other which implies that p p divides exactly one of the two factors. Therefore either 2 ( p 1 ) / 2 1 2^{(p-1)/2}-1 = = p x 2 px^2 , 2 ( p 1 ) / 2 + 1 2^{(p-1)/2}+1 = = y 2 y^2 or, 2 ( p 1 ) / 2 1 2^{(p-1)/2}-1 = = x 2 x^2 , 2 ( p 1 ) / 2 + 1 2^{(p-1)/2}+1 = = p y 2 py^2 CASE1:- Suppose 2 ( p 1 ) / 2 1 2^{(p-1)/2}-1 = = p x 2 px^2 , 2 ( p 1 ) / 2 + 1 2^{(p-1)/2}+1 = = y 2 y^2 . But then 2 ( p 1 ) / 2 2^{(p-1)/2} = = ( y 1 ) (y-1) ( y + 1 ) (y+1) for which the only solution is y y = = 3 3 i.e. p p = = 7 7 . When p p = = 7 7 we observe that 2 p 1 p \frac{2^{p-1}}{p} indeed a perfect square. CASE2:- Suppose 2 ( p 1 ) / 2 1 2^{(p-1)/2}-1 = = x 2 x^2 , 2 ( p 1 ) / 2 + 1 2^{(p-1)/2}+1 = = p y 2 py^2 . Then if p p > > 3 3 we get that x 2 x^2 leaves remainder 3 when divided by 4 which is impossible. Thus p p can only equal to 3 3 , when p p = = 3 3 also we observe that 2 p 1 p \frac{2^{p-1}}{p} indeed a perfect square. Thus the primes p p for which 2 p 1 p \frac{2^{p-1}}{p} is a perfect square are p = 3 , 7 p = 3 , 7

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