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Please help me to prove that any polynomial with integer coefficients can be represented as a sum of two irreducible polynomials over the ring $\mathbb{Z}$.

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    $\begingroup$ Isn't this a sort of Goldbach's conjecture for $\Bbb Z[x]$? $\endgroup$
    – ajotatxe
    May 3 '15 at 13:07
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    $\begingroup$ This was done by David Hayes, I believe in a short paper in the Amer. Math Monthly. See also math.stackexchange.com/questions/917092/…. $\endgroup$
    – KCd
    May 3 '15 at 13:22
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Let $$f(x)=a_0+a_1x+\ldots +a_nx^n$$ Let $p,q$ be two distinct odd primes. Then because $\gcd(p,q)=1$ we find $u_k,v_k$ such that $a_k=u_kp+v_kq$ and thus letting $$ g(x)=(u_0+rq)p+u_1px+\ldots +u_npx^n+x^{n+1}\\ h(x)=(v_0-rp)q+v_1qx+\ldots +v_nqx^n-x^{n+1}\\$$ (for some integer $r$) we have clearly that $f(x)=g(x)+h(x)$. Also, $g$ and $h$ do almost satisfy Eisentsein's irreducibility criterion: all coefficients except the leading one are divisible by $p$ (resp. $q$). However, we want to ensure that the constant term is not divisible by $p^2$ (resp. $q^2$), i.e., we need $p\nmid u_0+rq$ and $q\nmid v_0-rp$). But this can easily be achieved: At most one of the numbers $u_0-q, u_0,u_0+q$ is a multiple of $p$ (because the $\gcd$ of two of them divies $2q$) and at most one of $v_0+p,v_0,v_0-p$ is a multiple of $q$. Hence at least one of the choices $r\in\{-1,0,1\}$ leads to $p^2\nmid u_0+rq$ and $q\nmid v_0-rp$. With this choice, $g$ and $h$ are irreducible per Eisenstein.

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    $\begingroup$ You should say that this is essentially the proof by Hayes. $\endgroup$
    – KCd
    May 3 '15 at 13:41
  • $\begingroup$ @KCd I could not because I didn't know that. This proof was just the first that came to my mind. - Actually, the link in the answer here is dead and it appears from what Dietrich Bured wrote there, that Hayes did a little better by keeping degree $n$ whereas I took the freedom to increase the degree (which maybe makes the argument simpler and allows to catch the degree $0$ case) $\endgroup$ May 3 '15 at 13:46
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    $\begingroup$ Fair enough. I should have just said that this is essentially the proof of Hayes. $\endgroup$
    – KCd
    May 3 '15 at 14:23

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