How to use Legendre symbol to find a prime which divides $ax^2+b$?

Question

I'm trying to prove that $\dfrac{x^2-2}{2y^2+3}$is never an integer if $x,y\in\mathbb{Z}$.

It can be proven if $\forall p\in\mathbb{P}\:$doesn't suffice both of the following congruences: $$\: x^2\equiv 2 \pmod{p}$$ $$2y^2\equiv -3\pmod{p}$$

If I could know that which prime $p$ suffices the congruence $ax^2+b\equiv 0\pmod{p}$ where $a\not\equiv 0\pmod{p}$ (in this case, $2y^2\equiv -3\pmod{p}$), I can prove the claim.

Could you give me any idea?

Answer 1

There are odd primes $p$ such that the congruences $x^2\equiv 2\pmod{p}$ and $2y^2\equiv -3\pmod{p}$ have solutions. For example, $p=7$ works ($x=y=3$).

OP has asked for which (odd) primes $p$ does the congruence $2y^2\equiv -3\pmod{p}$ have a solution. It is not difficult to show that this congruence has a solution if and only if the congruence $z^2\equiv -6\pmod{p}$ has a solution.

Let $p\gt 3$. We want to find the primes such that the Legendre symbol $(-6/p)$ is equal to $1$. Since $(-6/p)=(2/p)(-3/p)$, there are two cases to consider, $(2/p)=1$ and $(2/p)=-1$.

In the first case, we want $(-3/p)=1$, and in the second case we want $(-3/p)=-1$.

We deal with the first case only. The analysis of the second case is very similar.

Because $(2/p)=1$, we want $p\equiv \pm 1\pmod{8}$. We have $(-3/p)=1$ if and only if $p\equiv 1\pmod{6}$, that is, $p\equiv 1$ or $7$ or $13$ or $19$ modulo $24$. The condition $p\equiv \pm 1\pmod{8}$ rule out the last two. So the first case holds if $p\equiv 1\pmod{24}$ or $p\equiv 7\pmod{24}$.

Remark: Finding for which $p$ we have $(-3/p)=1$ is straightforward. We divide into $p\equiv 1\pmod{4}$, which makes $(-1/p)=1$, and $p\equiv 3\pmod{4}$, which makes $(-1/p)=-1$.

For the case $p\equiv 1\pmod{4}$, we get $(3/p)=(p/3)$, so we want $p\equiv 1\pmod{3}$. In the case $p\equiv 3\pmod{4}$, we have $(3/p)=-(p/3)$, so again we end up wanting $p\equiv 1\pmod{3}$.

Answer 2

$$\frac{x^2-2}{2y^2+3}.$$

The primes that may possibly divide $x^2 - 2$ are $p=2$ and any $p \equiv \pm 1 \pmod 8.$ In preparation for the next part, these expand to $$ p = 2, \mbox{or} \; \; p \equiv 1,7,17,23 \pmod {24} $$

The primes that may possibly divide $2 y^2 +3$ are $$ p = 3, \mbox{or} \; \; p \equiv 1,5,7,11 \pmod {24}. $$ However, there is a fact you do not see very often, which is that the total exponent sum of primes $2,3$ or $\equiv 5,11 \pmod {24 }$ dividing $2y^2 + 3$ must be odd. This is Gauss composition.

As Andre points out, this means that $ \gcd(x^2 - 2, 2y^2 + 3) $ is the product of a bunch of primes $\equiv 1,7 \pmod {24}.$ BUT, if $2 y^2 + 3$ is divisible by $3,$ the ratio is not an integer. So, we may assume that $2 y^2 + 3$ is not divisible by $3.$ Therefore it is divisible by some prime $p \equiv 5,11 \pmod {24}.$ But $x^2-2$ is not divisible by that prime $p. \bigcirc \bigcirc \bigcirc \bigcirc$

Alright, so you need to know Gauss composition, genus theory, and one more fact: if $$ f(x,y) = a x^2 + b x y + c y^2, $$ then $$ \Delta = b^2 - 4 a c, $$ AND we have an odd prime $q$ such that $$ (\Delta | q) = -1, $$ and IF $$ a x^2 + b x y + c y^2 \equiv 0 \pmod q, $$ THEN $$ x,y \equiv 0 \pmod q $$ also.

Below is $2 y^2 + 3$ factored when $y$ is not divisible by$3,$ for small $y.$ Note the prime factors $5,11,29,53,59,83,101,107$ and so on.

  y                2 y^2 + 3
  1                   5 = 5
  2                  11 = 11
  4                  35 = 5 * 7
  5                  53 = 53
  7                 101 = 101
  8                 131 = 131
 10                 203 = 7 * 29
 11                 245 = 5 * 7^2
 13                 341 = 11 * 31
 14                 395 = 5 * 79
 16                 515 = 5 * 103
 17                 581 = 7 * 83
 19                 725 = 5^2 * 29
 20                 803 = 11 * 73
 22                 971 = 971
 23                1061 = 1061
 25                1253 = 7 * 179
 26                1355 = 5 * 271
 28                1571 = 1571
 29                1685 = 5 * 337
 31                1925 = 5^2 * 7 * 11
 32                2051 = 7 * 293
 34                2315 = 5 * 463
 35                2453 = 11 * 223
 37                2741 = 2741
 38                2891 = 7^2 * 59
 40                3203 = 3203

Answer 3

We show that $2y^2+3$ cannot divide $x^2-2$ using almost no information about the prime divisors of $2y^2+3$. This is given as a separate answer, since it does not characterize the primes $p$ for which $2y^2\equiv -3\pmod{p}$ has a solution. Such a characterization was asked for by OP, and done in our other answer. But it is unnecessary. We now proceed to the proof.

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Note that if $y$ is even, then $2y^2+3\equiv 3\pmod{8}$, while if $y$ is odd then $2y^2+3\equiv 5\pmod{8}$.

So in either case, $2y^2+3$ must have an odd prime divisor $p$ which is not congruent to $\pm 1\pmod{8}$. For a product of numbers of the form $8k\pm 1$ is again of that form.

Such a prime $p$ cannot divide $x^2-2$, by the standard elementary result about primes that have $2$ as a quadratic residue. This completes the proof.