For which $n$ the polynomial $x^n + y^n$ will be irreducible over $\mathbb Z$?
I think when n is multiple of 2. Can you please correct me if I am wrong and explain why.
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What do you mean by irreducible? Whether it can be further broken down into factors? – Nilabro Saha Jan 19 '17 at 04:03
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See here: http://math.stackexchange.com/questions/695266/factoring-xn-yn-over-the-integers – Jan 19 '17 at 04:05
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yes..Do you know the answer? – anonymous Jan 19 '17 at 04:05
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Rohan I have seen already. I could not understand – anonymous Jan 19 '17 at 04:06
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Unless I misunderstand, $x^6 + y^6 = (x^2)^3 + (y^2)^3 = (x^2 + y^2)(x^4 - x^2y^2 + y^4)$ – pjs36 Jan 19 '17 at 04:08
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yesss...U got a counter example. so the answer will be when n is a power of 2. Is not it???? – anonymous Jan 19 '17 at 04:09
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1Yes, $n$ must be a power of $2$. @sani – Thomas Andrews Jan 19 '17 at 04:18
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@Thomas Ansrews Is it a necessary and sufficient condition for being irreducible of $x^n + y^n$ – anonymous Jan 19 '17 at 04:23
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Pretty sure it is sufficient, but I'm blanking on how to prove it, at least, without a little abstract algebra. – Thomas Andrews Jan 19 '17 at 04:28
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When $n$ is a power of two it is irreducible over $\Bbb{Q}$ (see MathChat's answer). But it would be more precise if you specified the field. Over $\Bbb{R}$ it is irreducible only when $n=1$ or $2$, over many other fields it is never irreducible. I guess without a specified field we should assume the rationals – Jyrki Lahtonen Jan 19 '17 at 05:07
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Note that if m is odd, then $x^m+y^m$ is divisible by $x+y$. Thus for $n=2^r⋅m$, where $m>1$ is odd, it follows that $x^n+y^n=(x^{2^r})^m+(y^{2^r})^m$ is divisible by $x^{2^r}+y^{2^r}.$
On the other hand, polynomials of type $x^{2^r}+y^{2^r}$ are always irreducible over $\mathbb{Q}$. Note that any factorization of $f(x,y)=x^{2^r}+y^{2^r}$ will give us a factorization of $f(x,1)=x^{2^r}+1$. So it suffices to show that $x^{2^r}+1$ is irreducible. Recall that cyclotomic polynomials $$\Phi_{n}(x)=\prod_{\substack{1\leq k\leq n \\ (n,k)=1}} (x-e^{2\pi ik/n})$$ are irreducible over $\mathbb{Q}$. Now since
$$\Phi_{2^{r+1}}(x)=x^{2^r}+1,$$ the claim follows.
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