Prime One Less Than A Square

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

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,"

To address Melody's question:

2²ⁿ = 4ⁿ = (3 + 1)ⁿ

Expand

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

So 2²ⁿ - 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 2²ⁿ - 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.

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It doesn't state in the original question that the number has to be a power of 2.

It has to be even, so let it be 2m.

Squared and with 1 subtracted, that gets you 4m^2 - 1, and that factorises as (2m - 1)(2m + 1), which is composite unless m = 1.

Thank you Bertie and Alan, I really liked your explanations.