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I have to prove that the congruence $2x^2 +1 ≡ 0\pmod p$ has a solution if and only if $p ≡ 1 \text{ or } 3\pmod 8$, for any odd prime.

Thanks.

Marc van Leeuwen
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noddy
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  • Similar question: http://math.stackexchange.com/questions/333587/if-x4-equiv-1-mod-p-then-p-equiv-1-mod-8 – Inceptio Mar 18 '13 at 13:23

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You want $-2$ to be a square modulo $p$. In fact a solution $x$ exists if and only if $ x^{2} \equiv (-2)/4$ is a square modulo $p$, and thus if and only if $-2$ is a square modulo $p$.

This is because the following congruences modulo $p$ are equivalent: (reworked avoiding inverses at a suggestion of Math Gems in the comments)

  • $2 x^2 + 1 \equiv 0$
  • $2 x^{2} \equiv -1$
  • $(2 x)^{2} \equiv - 2$

Now $$ \left(\frac{-2}{p}\right) = \left(\frac{-1}{p}\right) \left(\frac{2}{p}\right) $$ is $1$ if and only if $p \equiv 1, 3 \pmod{8}$.

Andreas Caranti
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  • The first part is what I wanted to know how is 2x^2+1=0(mod p) same as x^2 = -2 (mod p) – noddy Mar 18 '13 at 13:18
  • @noddy, is it clear now? $2 x^2 + 1 \equiv 0$ is the same as $2 x^{2} \equiv -1$ which is the same as $x^{2} \equiv - 2^{-1} \equiv (-2) (2^{-1})^{2}$, where $2^{-1}$ denotes the inverse of $2$ modulo the odd prime $p$. – Andreas Caranti Mar 18 '13 at 13:20
  • yah..had initially missed the inverse part..Thanks – noddy Mar 18 '13 at 13:22
  • @noddy, you're welcome. – Andreas Caranti Mar 18 '13 at 13:24
  • infact using the technique even k*x^2+1=0(mod p) can be solved in a similar way;k would be invertible since p is a prime – noddy Mar 18 '13 at 13:29
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    @noddy $\rm\ mod\ p!:\ -1 \equiv kx^2:\Rightarrow -k\equiv k^2 x^2 \equiv (kx)^2.:$ Note that one does not need to use inverses to make that inference. It is squareness, not invertibility, that plays the key role. – Math Gems Mar 18 '13 at 13:45
  • @MathGems, quite right, I have edited accordingly, with proper credit. – Andreas Caranti Mar 18 '13 at 13:46