Question

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Answer 1

Answer:

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Step-by-step explanation:

Find all primes p such that p+1 is a perfect square.

All primes except for 2 (3 is not a perfect square, so we can exclude that case) are odd, so we can express them as 2n+1 for some n∈Z+. Let's express the perfect square as a2, where a∈Z+. Since we are interested in a number that is one more than 2n+1, we know that our perfect square can also be expressed as 2n+1+1=2(n+1).

2n+2=a2

2(n+1)=a2

So we know that our perfect square must be even, as it has a factor of 2 in it (in fact 2⋅2).

It is my strong intuition that we get a perfect square only if n=1, and therefore p=3 and p+1=a2=2⋅2=4, but how should I continue with this proof? It seems to me that whatever factor we have on the LHS we need to have it twice on the RHS (since a must be an integer),