Given equation has no integral solution.

Question

This question already has an answer here:

Diophantine equations to degree $2$

(1 answer)

Closed 2 years ago.

I need to show that $15x^2 – 7y^2 = 9$ has no integral solution.

Only thing that comes to my mind is to use to some modulo reduction here

I really have no idea how to proceed. Any hints????

Answer 1

Let us study $15x^2-7y^2=9$ in modular arithmetic:

In modulo $3$: $2y^2=0\rightarrow y^2=0\rightarrow y=0$. Hence $y=3z$ for some integer $z$. The new equation is $15x^2-63y^2=9\rightarrow 5x^2-21z^2=3$.

In modulo $3$ again: $2x^2=0\rightarrow x^2=0\rightarrow x=0$. Hence $x=3k$ for some integer $k$. The new equation is $45k^2-21y^2=3\rightarrow 15k^2-7z^2=1$.

Modulo $3$ one more time: $2z^2=1 \rightarrow z^2=2$. But notice that in modulo $3$ we have $0^2=0$, $1^2=1$ and $2^2=1$, so there is no integer $z$ such that $z^2=2$, thus the equation has no solution.