1

Let $p$ be a prime number, $p \geq 3$. Show that there exists a unique value of $p$ such that the polynomial $X^2 + X +1 $ is equal to $(aX^2 + b)^2$ for some values of $a,b$ in $\mathbb{F}_p$.

My attempt: we have $X^2 + X + 1 = a^2 X + 2ab X + b^2$, that means that $a^2 = 1, 2ab = 1, b^2 = 1$. And as we are working in $\mathbb{F}_p$, we can deduce that $$ a^2 \equiv 1 [p] , 2ab \equiv 1 [p], b^2 \equiv 1 [p]$$ Thus, $p \nmid a,b$, and so we have by the Fermat's theorem: $$a^{p-1}\equiv 1[p] $$ and $$b^{p-1}\equiv 1[p] $$ Thus $p = 3$.

Is my reasoning correct?

user26857
  • 52,094
John Mayne
  • 2,148
  • No. If it was correct, all that it would prove was that $3$ is a solution, not that it is the only solution. But you did not even prove that $3$ is a solution. If $a=b=1$, then $2ab=2\not\equiv1\pmod3$. – José Carlos Santos Dec 03 '17 at 09:46
  • 3
    Exactly how did you arrive at $p=3$??? Anyway, here's a hint. If such $a,b$ exist, then the polynomial $Q$ has a double root. Meaning that its discriminant is equal to .... – Jyrki Lahtonen Dec 03 '17 at 09:49
  • You also know that $x^2+x+1$ then would be reducible over $\mathbb{F}_p$ for $p>3$, hence that $p\equiv 1\bmod 3$, see here. – Dietrich Burde Dec 03 '17 at 09:51

1 Answers1

4

Jyrki's comment is probably the best way to go, but you could also argue from $a^2\equiv1\bmod p$ that $a\equiv\pm1\bmod p$, similarly from $b^2\equiv1\bmod p$ that $b\equiv\pm1\bmod p$; then these two imply $2ab\equiv\pm2\bmod p$, so $\pm2\equiv1\bmod p$. $2\equiv1$ goes nowhere, but $-2\equiv1$ gets you to the necessary condition on $p$ real fast.

Then you just have to check that $p=3$ works, but you've got the $a$ and $b$ from the work you've done already.

Gerry Myerson
  • 179,216
  • You reasoning work fine with me. But one of the comments on my questions states that if $p=3$ was the solution, then $a=b=1$ then $2ab = 2 \not \equiv 1 [3]$. I am unsure what is missing here. – John Mayne Dec 03 '17 at 10:02
  • That's not what the comment says. – Gerry Myerson Dec 03 '17 at 10:04
  • I am confused. Doesn't his comment disprove that $p=3$? – John Mayne Dec 03 '17 at 10:16
  • 1
    How could his comment possibly disprove $p=3$, when you can find $a$ and $b$ that show that $p=3$ works? The comment nowhere says "If $p=3$, then ...." All it says is you can't have $a=b=1$. – Gerry Myerson Dec 03 '17 at 10:56