Find the number of integer solutions of $22x^2-34xy+6y^2=2024$

Question

I was given this and I could really use some hints. I know that there are no integer solutions to the equation and that I have to factor the left side to get the form $(ax+by)^2$ by multiplying/dividing both sides of the equation and/or perhaps use a prime modulo of some sort so that I get a congruence in the form of $w^2\equiv c$ and then check that c is not a quadratic residue of the modulo and so reach the conclusion that there are no integer solutions to the problem. I have tried modules 3-37 (maybe made some mistakes along the way) and tried diving by 2 and then adding the modulo and haven't gotten even close to the right answer. Perhaps someone could push me toward the right modulo or give me a hint of what to look for.

Answer 1

Multiplying both sides by 6 gives you $(6y-17x)^2-157x^2=12144$.

But quadratic reciprocity gives you $\left(\dfrac{12144}{157}\right)=-1$, so this equation admits no integral solution.

Answer 2

One obvious minor simplification to the equation is to observe that all of the coefficients are even, so let's divide by 2.

$$11x^2-17xy+3y^2=1012$$

Let's try your modulo arithmetic approach, and use a computer program to help us.

def eqmod(a, b, c):
   '''
   Check if a is congruent to b modulo c.
   '''
   return (a - b) % c == 0

def solvemod(m):
   '''
   Return a set of (x, y) pairs that satisfy the equation
   11x^2 - 17xy + 3y^2 = 1012, modulo m.
   '''
   return {
       (x, y)
       for x in range(m)
       for y in range(m)
       if eqmod(11 * x ** 2 - 17 * x * y + 3 * y ** 2, 1012, m)
   }

I went ahead and computed all of the solutions sets for moduli up to 1000.

It turns out that if you pick $m = 157$ (or any multiple of 157), then there are no possible solutions. And $m = 529 = 23^2$ has no solutions either, despite not being a multiple of 157. So if you're willing to accept a brute-force computer proof, just do the arithmetic modulo 157 or 529.

Alternatively, we could use the largest modulus for which there is exactly one solution using modular arithmetic. That works out to $m = 46$, which has the solution $x = y = 0$. IOW, $x = 46p$ and $y = 46q$, where $p, q \in \mathbb{Z}$. Making this substitution gives:

$$11(46p)^2 - 17(46p)(46q) + 3(46q)^2 = 1012$$ $$46^2(11p^2 - 17pq + 3q^2) = 1012$$ $$11p^2 - 17pq + 3q^2 = \frac{11}{23}$$

So we have an integer on one side of the equation, and an non-integer on the other. That doesn't work.