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Given $z^2=10x^2-5y^2$.

I am wanting to show this equation has integer solutions only $0=x=y=z$.

I attempted by considering modulo $2$, however this didn't give me a contradiction. Any help or hints would be appreciated.

Solution:

Consider modulo $5$. Then we have that $z^2 \equiv 0 \mod5$. This is impossible as:

$0^2=0 \mod5$

$1^2=1 \mod5$

$2^2=4 \mod5$

$3^2=4 \mod5$

$4^2=1 \mod5$

Asaf Karagila
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babynewton
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    https://math.stackexchange.com/questions/2773097/how-to-find-all-rational-solutions-of-x2-3y2-7/2788381#2788381 – individ Jul 26 '19 at 04:23
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    Try modulo 5 instead. Do it recursively for LHS and RHS. – Hoseyn Heydari Jul 26 '19 at 04:32
  • Looking modulo $4,$ if there is a non-trivial solution, there must be one with $x$ odd. – Thomas Andrews Jul 26 '19 at 04:39
  • @HoseynHeydari I have edited the above to include a solution, is this sufficient for a proof? – babynewton Jul 26 '19 at 04:53
  • Why is it impossible? You have shown that $z \equiv 0 \pmod 5$. That is a good step. Now what about $x,y$? Maybe they are multiples of $5$ as well. You should be able to use your table of squares to show they must be. – Ross Millikan Jul 26 '19 at 04:57
  • @babynewton The right hand side of your equation is congruent to $0$ modulo $5$, so I don't quite know how you can state, with what you've said, that it's impossible. Note it is, but for a somewhat different reason, which Hoseyn Heydari's comment indicated, and I've explained further in my answer. – John Omielan Jul 26 '19 at 04:57

2 Answers2

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As Hoseyn Heydari's comment stated, you should consider checking modulo $5$ instead for your equation of

$$z^2 = 10x^2 - 5y^2 \tag{1}\label{eq1}$$

As the RHS is a multiple of $5$, this means $z = 5z_1$ for some integer $z_1$. Thus, \eqref{eq1} becomes

$$5z^2_1 = 2x^2 - y^2 \tag{2}\label{eq2}$$

Since the possible congruences of squared values modulo $5$ are $0, 1$ and $4$, then $2x^2$'s values modulo $5$ must be $0, 2$ or $3$ (as $2 \times 4 \equiv 3 \pmod 5$). Thus, since the LHS is a multiple of $5$ and the only match between the $2$ sets of $\{0,2,3\}$ and $\{0,1,4\}$ is $0$, this means $2x^2 - y^2$ can only be congruent to $0$ modulo $5$ if both $x$ and $y$ are as well. Let $x = 5x_1$ and $y = 5y_1$ to get

$$z^2_1 = 10x^2_1 - 5y^2_1 \tag{3}\label{eq3}$$

Note this is of the same form as \eqref{eq1}, but with each value being $5$ times smaller. As the procedure can be repeated indefinitely, this is only possible if each of the values were $0$ initially, giving that the sole solution is $x = y = z = 0$.

John Omielan
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  • Alternative argument: $2x^2\equiv y^2\mod5\implies x\equiv y\equiv0\mod5$ because $2$ is not a quadratic residue $\mod 5$ – J. W. Tanner Jul 26 '19 at 07:46
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Outline of proof:

  1. If there is a non-trivial solution, there is a solution with $x$ odd.
  2. There are no solutions when $x$ odd.

We'll look entirely modulo $8.$

First note that if $x,y,z$ are solutions with $x=0$, then $y=0,z=0.$

So if there is any solution $(x,y,z)\neq (0,0,0),$ there must be one with $x>0.$ Then find the least $x>0$ such that there are integers $y,z$ such that $z^2=10x^2-5y^2.$

If $x$ is even, then $10x^2\equiv 0\pmod{8}.$ We then have $z^2+5y^2\equiv 0\pmod 8.$

But $$z^2\equiv 0,1,\text{ or } 4\pmod 8\\\text{and}\\5y^2\equiv 0,5,\text{ or } 4\pmod{8}.\tag{1}$$ So to get $z^2+5y^2\equiv 0\pmod 8,$ we must have $y,z$ both even. Then we get another solution $(x',y',z')=(x/2,y/2,z/2),$ contradicting that $(x,y,z)$ is a solution with minimal $x>0.$

So $x$ must be odd.

When $x$ is odd, we have $10x^2\equiv 2\pmod{8}.$

By (1) we get that $$z^2+5y^2\equiv 0,1,4,5,\text{ or }6\pmod{8}$$

So it is not possible for $z^2+5y^2\equiv 2\pmod{8}.$ But that means there is no $(x,y,z)\neq 0$ such that $z^2=10x^2-5y^2.$

More generally, if $a\equiv 1\pmod{8},$ $b\equiv 5\pmod 8$ and $c\equiv 2,3,7\pmod 8$ then there are no integer solutions $(x,y,z)$ with $x\neq 0$ to the equation:

$$az^2+by^2=cx^2.$$

Thomas Andrews
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