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I am trying to determine whether the equation below has a solution or not

$$x^2-97y-40 =0.$$

If a solution exists, $x^2-40$ must be congruent to $0$ modulo $97$.

If I could show the congruence above implies that solution exists.

Thanks for your help...

Martin Sleziak
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user108752
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    The question is whether 40 is a quadratic residue modulo 97. Methods for determining this are given in virtually every introductory number theory textbook ever written. I'd encourage you to borrow one from your nearest library, and read up on it. Or just search the web for "quadratic residue", "quadratic reciprocity", "quadratic congruence". – Gerry Myerson Nov 14 '13 at 12:11

2 Answers2

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$\newcommand{\kron}[2]{\left( \frac{#1}{#2} \right)}$

We are interested in the value of the Legendre symbol $\kron{40}{97} = \kron{8}{97} \kron{5}{97}$. In general, $2$ behaves differently from the rest of the primes.

By what is sometimes called the second supplemental law of quadratic reciprocity, we know that $\kron{2}{p} = (-1)^{(p^2-1)/8}$, or rather that $2$ is a square mod a prime iff that prime is congruent to $\pm 1 \pmod 8$. Since $97 \equiv 1 \pmod 8$, we see that $2$ is a square mod $97$. (Proofs of the second supplemental law can be found both here on MSE and on my site, and in most elementary number theory texts).

We must still consider $\kron{5}{97}$. By quadratic reciprocity, this is the same as $\kron{97}{5} = \kron{2}{5}$. The squares mod $5$ are $1$ and $4$ (or alternatively, $5$ is not congruent to $\pm 1 \pmod 8$), so $\kron{2}{5} = -1$.

In total, we have that $\kron{40}{97} = 1 \cdot -1 = -1$, which means that $40$ is not a square mod $97$. And so this quadratic Diophantine equation has no solution. $\diamondsuit$

Community
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davidlowryduda
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you can use legendre symbol for calculating quadratic residue.so you need to compute (40/97).

  (40/97)=(2/97).(2/97).(2/97).(5/97)

        =(2/97).(5/97)

        =(97/2).(97/5)

        =(1/2).(2/5)

       =1.(-1)=-1

so 40 is not a quadratic residue mod 97.

no solution exists.

juliet
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    $(2/97)=(97/2)$ is an invalid argument, unless you provide some justification --- e.g., it's not true that $(2/5)=(5/2)$. – Gerry Myerson Nov 17 '13 at 12:17