Find the sum of all the prime numbers for which
is a perfect square
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
For p = 2 the p 2 p − 1 is not an even integer and so we can assume p is an odd prime. Then by Fermat's Little Theorem 2 p − 1 − 1 is divisible by p . Suppose that for some integer a , 2 p − 1 − 1 = p a 2 .Since p is odd we have that ( 2 ( p − 1 ) / 2 − 1 ) ( 2 ( p − 1 ) / 2 + 1 ) = p a 2 Both the factors on the L.H.S are odd and hence they are relatively prime to each other which implies that p divides exactly one of the two factors. Therefore either 2 ( p − 1 ) / 2 − 1 = p x 2 , 2 ( p − 1 ) / 2 + 1 = y 2 or, 2 ( p − 1 ) / 2 − 1 = x 2 , 2 ( p − 1 ) / 2 + 1 = p y 2 CASE1:- Suppose 2 ( p − 1 ) / 2 − 1 = p x 2 , 2 ( p − 1 ) / 2 + 1 = y 2 . But then 2 ( p − 1 ) / 2 = ( y − 1 ) ( y + 1 ) for which the only solution is y = 3 i.e. p = 7 . When p = 7 we observe that p 2 p − 1 indeed a perfect square. CASE2:- Suppose 2 ( p − 1 ) / 2 − 1 = x 2 , 2 ( p − 1 ) / 2 + 1 = p y 2 . Then if p > 3 we get that x 2 leaves remainder 3 when divided by 4 which is impossible. Thus p can only equal to 3 , when p = 3 also we observe that p 2 p − 1 indeed a perfect square. Thus the primes p for which p 2 p − 1 is a perfect square are p = 3 , 7