# Prove that there are no non-trivial integer solutions to the Diophantine equation.

$$3x^2 - 5y^2 + 7z^2 = 0$$

I am totally lost with this question and I have no clue where or even how to start this. There is a little hint given saying I might want to consider the equation modulo different integers. I don't even know how to do modulos for I was never taught.

• If you have not yet seen things like $k \pmod m$ then this problem is probably too hard for you. But you can use the hint by trying things like "Let $x = 13 s + X, y = 13 t + Y, z = 13 u + Z$ with $X,Y,Z$ in $[0,12]$ and seeing what happens when you square $x$ and $y$ and $z$. I'm not saying that the $13$ is the right number to try, but this is the hint idea. – Mark Fischler Oct 12 '16 at 3:53
• Why did you choose 13? I do have some minimal knowledge with modulos but not enough to know what to do. – Chance Gordon Oct 12 '16 at 4:17

The first step in a problem of this sort is usually to say that if a solution exists, then some solution exists in which the greatest common divisor of $x, y, z$ is 1. For if $x=px', y=py', z=pz'$ then you can get a smaller but proportional triplet by factoring out $p^2$ to get $$3x'^2 -5y'^2 +7z'^2 = 0$$

Now that you know they are not all divisible by the same number, so that in modulo $m$ the solution $(0,0,0)$ is not allowed, you can start looking at various mods. For a problem with three variables, it is always tempting to try a modulus that will make one of the terms disappear identically -- $3, 5$ or $7$ in this case. Let's try $3$: Write $$x = 3a + b \\ y = 3c + d \\ z = 3e+f$$ with $b,d,f \in \{0,1,2\}$. Then the original equation becomes $$3(9a^2+6ab+b^2) - 5 (9c^2 +6cd+d^2) + 7(9e^2 +6ef+f^2) = 0$$ And then even if you never heard the word mod, you can do $$3(9a^2+6ab-15c^2-10cd+21e^2+14ef + b^2 -6d^2 +2f^2) + d^2 + f^2 = 0$$

So $d^2+f^2$ has to be divisible by $3$; but the only way to do that with $d$ and $f$ in $\{1,2,3\}$ is to have $d=f=0$. So $y = 3c$ and $z=3e$, and $x$ had better not be divisible by $3$, otherwise the g.c.d. is not $1$.

Then the original equation becomes $$3x^2 -45c^2 + 63e^2 = 0 \implies x^2 -15c^2 + 21e^2 = 0 \implies x^2 = 3(5c^2-7e^2)$$ So try as hard as we might, we have to take $x^2$ divisible by $3$, and so the triplet, if it exists, cannot have a g.c.d. of $1$. So no such triplet can exist.

• (+1) A very nice and clear explanation which might even be ok to follow if you don't understand mod-operations. Two small comments: (1) the first equation reads $-7z'^2$ which should be $+7z'^2$, and (2) you wrote "$d$ and $f$ in $\{1,2,3\}$" where you meant "$d$ and $f$ in $\{0,1,2\}$". – TMM Oct 14 '16 at 4:41
• thank @TMM. Changes made. – Mark Fischler Oct 16 '16 at 17:26

Modulo basically means remainder. A number

$$n\mod m$$

(read $n$ modulo (or mod) $m$) is the remainder when $n$ is divided by $m$. Furthermore,

$$n\equiv k\mod m$$

(read $n$ is equivalent to $k$ mod $m$) means that $n$ and $k$ give the same remainder when divided by $m$, or that

$$m|(n-k)$$

This is useful because we can consider the "residues" (basically meaning remainders) that applying different functions to numbers yields, $\mod m$. For instance, the residues of $x^2\mod m$ are just the set of remainders that $x^2$ can leave when divided by $m$. When $m=4$, we have that

$$0^2\mod 4 = 0$$

$$1^2\mod 4 = 1$$

$$2^2\mod 4 = 0$$

$$3^2\mod 4 = 1$$

$$(4n+k)^2\equiv 16n^2+8nk+k^2 \equiv 4(4n^2+2nk)+k^2\equiv k^2\mod 4$$

Because of these relations, the residues that $x^2$ can leave $\mod 4$ are $0$ and $1$. From this, we know that there exist no integer solutions to

$$x^2=4k+2$$

or

$$x^2=4k+3$$

which can be very important in number theory.

Hint on this specific problem: Consider the problem $\mod 3$. What residues could $3x^2$,$-5y^2$, or $7z^2$ leave $\mod 3$? When does the equation equal $0\mod 3$?

• Why did you choose mod 3? – Chance Gordon Oct 12 '16 at 4:18
• @ChanceGordon there is no hard and fast rule. We check out the some simple mod, though we usually begin with the coefficients of the variables – N.S.JOHN Oct 12 '16 at 4:21
• @ChanceGordon To add on to what N.S.JOHN said, I tried each coefficient. $3$ works, so I stopped there. Usually you'll want to do the smallest modulus possible, so I began with $3$. – Carl Schildkraut Oct 12 '16 at 4:34
• Ok, so should I start by saying 3x^2 − 5y^2 + 7z^2= 0 mod 3? – Chance Gordon Oct 12 '16 at 4:43
• I'm a bit surprised you give the example modulo $4$, showing that the residues are $0,1$, and then you don't follow up on this for this specific problem. If I am not mistaken, modulo $4$ the equation translates to $x^2 + y^2 + z^2 \equiv 0$, which has only one solution $x^2 \equiv y^2 \equiv z^2 \equiv 0$, in which case $\gcd(x,y,z) \geq 2$. Am I missing something? – TMM Oct 14 '16 at 4:35