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Prove that if $a$ is any integer and the polynomial $f(x) = x^2 +ax+ 1$ factors (poly mod 9), then there are THREE distinct non-negative integers $y$ less than $9$ such that $f(y) ≡ 0 (\bmod 9)$.

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Hint $\ $ mod $\,9,\,$ the discriminant is a square iff it is $\equiv 0\,$ iff $\,f(x) = (x\!-\!a)^2,\,$ and then

$$ 9\mid (x\!-\!a)^2\iff 3\mid x\!-\!a\iff x\equiv a,\,a\!+\!3,\,a\!+\!6\!\!\pmod{9}$$

Alternatively: the only factorizations mod $3$ are $\,(x\pm1)^2.\,$ Examine how those lift mod $9$.

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  • $\begingroup$ Hey okay, I understand. But now what do I do $\endgroup$ – Nadine Feb 13 '15 at 21:26
  • $\begingroup$ @Miss We're done, since the above shows there are exactly $3$ roots $\,a,\, a\!+\!3,\, a\!+\!6\pmod{9}\ \ $ $\endgroup$ – Bill Dubuque Feb 13 '15 at 21:28
  • $\begingroup$ Okay. So I'll be fine with just that? $\endgroup$ – Nadine Feb 13 '15 at 21:39
  • $\begingroup$ What do you mean by lift mod 9 $\endgroup$ – Nadine Feb 13 '15 at 22:02

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