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Show that the equation $x^2+y^2+z^2=7 w^2$ has no non-trivial solutions in integers.

This is a statement made in Lam's Introduction to Quadratic Forms over Fields (Chpt 1, Sec 2). "$7$ is known to be not in $D(f)$ in elementary number theory" where $D(f)=\{(x,y,z)\in Q^3, x^2+y^2+z^2=7\}$ and $Q$ is rational number.

It is easy by exhaustion to check $\mod(8)$ admitting no solution for $(x,y,z)$ which I checked by Mathematica. Thus there are no integral solution.

$\textbf{Q:}$ Now I want to check that there is no integer solution. In other words, I need to check $x^2+y^2+z^2=7w^2$ with $w\in \mathbb{Z}-\{0\}$. How do I check this? I tried mod 8 but it seems that this does not say anything about non-existence.

darij grinberg
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user45765
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  • see http://alpha.math.uga.edu/~pete/Clark_Jagy_11_13_2013.pdf – Will Jagy Feb 27 '19 at 20:06
  • ...Just clear denominators? If you had a rational solution, doing this gives you an integral solution which you proved does not exist. – YiFan Tey Apr 25 '19 at 03:22
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    I think mod 8 works, you have $x^2+y^2+z^2+w^2=0\mod 8$ and you can assume $\gcd(x,y,z,w)=1$. I don't think there are solutions. – Julian Mejia Apr 25 '19 at 03:28
  • @JulianMejia Unfortunately, I have already tried mathematica to look up the table of mod 8 and there are non-trivial points in $Z_8^4$ giving rise to mod $8$ solutions. That is why I am asking the question. – user45765 Apr 25 '19 at 03:30
  • Can you tell me which are these solutions? These solutions should happen only when $x,y,z,w$ are even, but we assumed coprimility. – Julian Mejia Apr 25 '19 at 03:34
  • @JulianMejia Sorry. I think I forgot to drop coprimeness condition. So after dropping coprimeness condition. Then it follows that there are no solutions. May I ask for a heuristic explanation for why one expects solutions happening only when x,y,z,w are even? – user45765 Apr 25 '19 at 03:39
  • I just wrote the solution, I am using that $(odd)^2 =1\mod 8$ – Julian Mejia Apr 25 '19 at 03:46
  • https://math.stackexchange.com/questions/1127654/parametrization-of-solutions-of-diophantine-equation – individ Apr 25 '19 at 04:17

2 Answers2

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Let's say that $(x,y,z,w)$ is a solution. By dividing by $\gcd(x,y,z,w)$ we can assume $\gcd(x,y,z,w)=1$.

We have $x^2+y^2+z^2+w^2=8w^2$ By looking mod 2, we have only two options, that two of them are even or that all of them are odd.

1st case: WLOG say $x,y$ even and $z,w$ are odd, then $x^2+y^2+z^2+w^2=0+0+1+1 \mod 4$ giving a contradiction.

2nd case: If $x,y,z,w$ are odd, then $x^2+y^2+z^2+w^2=1+1+1+1 \mod 8$ giving a contradiction as well.

So, there are no non trivial solutions.

Julian Mejia
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Hint: We are looking for integers $a,b,c,d$ with $a^2+b^2+c^2 = 7d^2$. Work modulo a power of $2$.

We may assume $a,b,c,d$ are coprime. The only quadratic residues modulo $8$ are $0, 1, 4$ so the LHS is $0, 1, 2, 3, 4, 5$ or $6$ mod $8$ and the RHS is $0, 4$ or $7$. So we must have that both sides are $0$ or $4$ mod $8$. But this can only happen when $a,b,c,d$ are divisible by $2$, which contradicts that they are coprime.

(Why does this work? by Hasse-Minkowski, there must be some number modulo which it has no nontrivial solutions. The obstruction comes from invariants such as the discriminant, $-7$. But working modulo $7$ doesn't help much. Since there are squares, and squares behave weird in characteristic $2$, it's reasonable to try working modulo powers of $2$. But to be honest, I wrote this hint directly with Legendre's 3-square theorem in mind.)

Bart Michels
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