Show that this polynomial in two variables has no integer solutions

Question

Consider the equation $15x^2 - 7y^2 = 16$. Show that it has no integer solutions.

Any hints or comments are welcome.

Answer 1

Hint : Considering in mod $m$ for some $m\ge 2\in\mathbb N$ sometimes works. In our case, $m=3$.

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Added : It seems that you are not familiar with considering in mod $m$.

The equation can be written as $$y^2=3(5x^2-2y^2-6)+2$$ So, $y^2$ has to be of the form $3k+2$ where $k\in\mathbb Z$. This is impossible. (why?)

Answer 2

Consider the equation modulo 3: $$15x^2-7y^2=16$$

• $15x^2$ is always divisible by 3,
• $7y^2$ is either 0 or 1 modulo 3 and
• $16\equiv1\pmod3$.

Substituting these residues into the equation, we find that it is inconsistent in all cases: $$0-0\not\equiv1\pmod3$$ $$0-1\not\equiv1\pmod3$$ $$15x^2-7y^2\not\equiv16\pmod3$$ Therefore the equation has no integer solution.

Answer 3

A good learning Software for your problem was written by Dario Alejandro Alpern:

set mode to Step-by-Step.

First of all we must determine the $gcd$ of all coefficients but the constant term, that is: $gcd(15, 0, -7, 0, 0) = 1.$

Dividing the equation by the greatest common divisor we obtain: $15 x^2 - 7 y^2 - 16 = 0$.

We try to check the equation modulo the prime divisors of $15$ ... here the hint ends :) ...

Answer 4

This may be a bit long. Assuming x, y are integers, they can either be even or odd. But only one of them being even and other being odd cannot be solution, since difference of odd and even numbers will not be even.

So they either have to both even or both odd.

If x and y are even then equation will be 60a-28b =16.

First term always ends in 0, second term has to end in 4. Meaning b which is a square should end in either 3 or 8, and there are no such numbers. So x cannot be even and y cannot be odd

Last possibility is x, y both are odd.

If x is odd, then first term will always end in 5. And second term has to end in 9. Meaning y square should end in 7. Since there are no such squares, x,y both cannot be odd.

Thus the equation cannot have integer solution.

It's a bit long though.