\( \mathbf{Theorem:} \) every prime except \( 2 \) can be written as a difference of two square numbers
We have to find two integers and such that for a prime
now we can rewrite equation as,
now as is a prime number we can't write as a product of two integers except for and itself. from this, we get that,
solving equation and , we get and .
notice that and are integers if is an odd number. Since all primes except for are odd, our proof is complete for every odd primes.
now we have to proof for the even primes, which is only . again notice that and can't be integer if is an even number. so we can't write as if is an even integer.
so from here, we can only deduce that we can write as for any prime except for
a friend of mine claims my proof is incomplete but I can't find it. can you point it where?
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If I'm not mistaken, your proof is backwards; that is, you started by assuming that every odd prime can be written as a difference of perfect squares, and then verifying that a and b are solutions. Essentially, what you have proven is that "if p=a2−b2, then a=(p+1)/2 and b=(p−1)/2 is a solution." This is not quite the same as, though very closely related to, proving "there exist a and b such that p=a2−b2." The quick fix to this problem would be to reverse your steps.
However, I believe you have successfully shown that 2 cannot be written in the form a2−b2 by contradiction. So, I would guess that your friend was referring to the above issue.
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Indeed. Even after we found a sequence of steps that lead to a solution/proof, we have to be careful with how it is written up, and ensure that the statements flow in the direction that we desire. It is very common to end up with implication in the opposite direction of what is desired.