I tried factoring it and got $(2x+1)(2x-1)$, however I do not know how to prove for all integers from here.
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Try also $4x^2+1$, if you have time. Again $x=1$ is a solution, but to find all $x$ is more difficult. – Dietrich Burde Nov 21 '13 at 20:10
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I have given it a try, I'm ashamed to say I cannot even factor that thing. – Bygonaut Nov 21 '13 at 20:25
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You are on a good way. Factoring does not work. – Dietrich Burde Nov 21 '13 at 20:28
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Well, you're trying to show that $4x^2-1$ is prime for some integer $x$, but you just factored it! If $4x^2-1$ is going to be prime, your factorization has to a trivial factorization. We can always factor primes as $p=p\cdot 1$, so maybe this will lead you to the correct answer.
JMag
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A prime number is one where its only factors are itself and one. The factoring you performed gives two factors, so to be prime one of these must be 1 or -1.
Solving $(2x+1)=1$ gives us $x=0$ and the other factor is $-1$. This isn't prime. Solving $(2x-1)=1$ gives us $x=1$ and the other factor is $3$. This is the only prime of that form.
Solving $(2x+1)=-1$ gives us $x=-1$ and the other factor is $3$, which works. Solving $(2x-1)=-1$ gives us $x=0$ again.
So the candidates are $x=1,-1$.
RussH
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Thank you @RussH and Dietrich, it was on the tip of my tongue, but I wasn't sure how to get there. Very clear explainations from both of you. – Bygonaut Nov 21 '13 at 20:16
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Good to hear! If you feel that the question has been answered, it is customary to mark it as 'answered'. – RussH Nov 25 '13 at 03:02