It can be seen that 2 squared - 1 =3 is a prime. Find the next example which is one less than a perfect square and is prime.

Guest Feb 27, 2015

#1**+10 **

The perfect square would have to be even and be of the form (2^{n})(2^{n}) = 2(^{2n})

And the prime number would be 2(^{2n}) - 1

But...it can be shown that for any such prime, the exponent 2n would have to be prime, and this is impossible for n > 1.

So....no such further examples exist......

CPhill
Feb 27, 2015

#1**+10 **

Best Answer

The perfect square would have to be even and be of the form (2^{n})(2^{n}) = 2(^{2n})

And the prime number would be 2(^{2n}) - 1

But...it can be shown that for any such prime, the exponent 2n would have to be prime, and this is impossible for n > 1.

So....no such further examples exist......

CPhill
Feb 27, 2015

#2**+5 **

that's interesting Chris

but

Would you like to show us how this property can be shown Chris.

"But...it can be shown that for any such prime, the exponent 2n would have to be prime,"

Melody
Feb 27, 2015

#3**+5 **

To address Melody's question:

2^{2n} = 4^{n} = (3 + 1)^{n}

Expand

(3 + 1)^{n} = 3^{n} + n×3^{n-1} + n(n-1)×3^{n-2}/2 + ... + n×3 + 1

So 2^{2n} - 1 = (3 + 1)^{n} - 1 = 3n + n×3^{n-1} + n(n-1)×3^{n-2}/2 + ... + n×3

Every term on the far right is exactly divisible by 3, hence 2^{2n} - 1 is exactly divisible by 3. The only prime number that can be exactly divisible by 3 is 3 itself. Hence there are no further examples of the type asked for in the original question.

.

Alan
Feb 27, 2015