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Can we represent any prime $p \equiv 1 \pmod{5}$ or $p \equiv -1 \pmod{5}$ in terms of $(a+b)^2 + ab$ with $a> b> 0$?

Can we represent any prime $p \equiv 1 \pmod{3}$ in terms of $(a+b)^2 - ab$ with $a> b> 0$?

can we have, any odd integer $> 1$ can be written as $a + b$ where $a$ and $b$ are positive integers with $a^4$ + $b^4$ is prime?

Kindly discuss. Thanks in advance

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The statement is not clear. It seems to be an incomplete sentence. Are you missing a second part to the statement? Also, what does $N$ have to do with anything? – Ted Nov 8 '12 at 4:13
@TED! N is > 1 and the sum of a + b is >1 being a & b are positive numbers. when p is congruent to + or - (mod 5), the prime p can be written as (a+b)^2 + ab uniquely. for a >0 and b >0. Similarly, can we do the same for my second part? – gama Nov 8 '12 at 4:38
Still not clear. Suddenly you have a condition on $p$ that you didn't originally have. Can you edit the questions to include all the conditions with the complete statements? – Ted Nov 8 '12 at 4:43
@Ted!I am poor in English.however, I made my post in more proper way. Now I am looking for your reply. Thank you for your support. – gama Nov 8 '12 at 5:04
I don't know what that means. – Gerry Myerson Nov 8 '12 at 6:08
up vote 3 down vote accepted

As another answer shows, if $p\equiv1\pmod3$, then $-3$ is a quadratic residue modulo $p$. That means there is an integer $x$ such that $$p{\rm\ divides\ }x^2+3$$ Now we pass to ${\cal O}_{-3}$, the ring of integers in the number field $K={\bf Q}(\sqrt{-3})$, and write, $$p{\rm\ divides\ }(x+\sqrt{-3})(x-\sqrt{-3})$$ Now $p$ doesn't divide either one of $x\pm\sqrt{-3}$ since $${x\over p}\pm{1\over p}\sqrt{-3}$$ are not in ${\cal O}_{-3}$. But it is known that ${\cal O}_{-3}$ is a unique factorization domain, implying that if an irreducible divides a product it must divide one of the factors. We deduce that $p$ is not an irreducible in ${\cal O}_{-3}$. Let $$\pi=a+b{1+\sqrt{-3}\over2}$$ be a nontrivial factor of $p$ in ${\cal O}_{-3}$. Then the norm of $\pi$ is a positive nontrivial factor of the norm of $p$, which is $p^2$, so the norm of $\pi$ is $p$. But the norm of $\pi$ is $a^2+ab+b^2$, QED.

EDIT: Similarly, if $p\equiv\pm1\pmod5$, then $5$ is a quadratic residue modulo $p$, so $$p{\rm\ divides\ }x^2-5$$ for some $x$. Going to ${\cal O}_5$, $$p{\rm\ divides\ }(x+\sqrt5)(x-\sqrt5)$$ but $p$ divides neither one of $x\pm\sqrt5$. Now ${\cal O}_5$ is known to be a UFD, so, again, $p$ is not irreducible in ${\cal O}_5$. Let $$\pi=r+s{1+\sqrt5\over2}$$ be a nontrivial factor of $p$ in ${\cal O}_5$. Taking norms, we get $$p=r^2+rs-s^2$$ Now a simple substitution gets us to $(a+b)^2+ab$.

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! what about the first and last questions? – gama Nov 9 '12 at 4:35
What about the question I asked you? Namely, what do you mean by "I am searching by computations in different ways"? – Gerry Myerson Nov 9 '12 at 4:50

I can answer for your second part. But, you should know about reciprocity. Otherwise, I am sorry. p = 1 (mod 3) is very known. for (-3/p) = (-1/p)(3/p) =) (-1/p) = (-1)^{(p-1)/2} and (3/p) = (-1)^{(p-1)/2}(p/3) =) (-3/p) = (-1)^{(p-1)/2}(-1)^{(p-1)/2}(p/3) = (p/3) i.e., -3 is quadratic residue mod p, if and only of p = 1 (mod 3). NOTE you can try for p = 2 (mod 3) more easily without using reciprocity. I hope you got it.

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! This time your solution is not impressed me. Try to answer first and last questions. – gama Nov 9 '12 at 4:35
! your NOTE is good. I will try to solve. – gama Nov 9 '12 at 5:49
! why can't we solve with reciprocity? – gama Nov 9 '12 at 5:52

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