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Given the equations: $$6x^2+y^2=z^2\\x^2+ 6y^2=t^2$$ for all $x,y,z,t \in\mathbb{N}$. Solve for $x, y, z$, and $t$.

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up vote 1 down vote accepted

The only solution is zero. Adding the two equations we get

$$7(x^2+y^2)=z^2+t^2$$ thus $$z^2+t^2\equiv 0 (mod 7)$$ and if $zt \neq 0$ we have that $-1$ is a quadratic residue modulo $7$, as $7\equiv 3 (mod 4)$ this is not the case so $xt=0$ and it follows that all are zero.

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Your answer is better and simple! – Sagnik Saha Jun 2 '14 at 15:33
Thanks, both proofs use the fact that $-1$ is not a quadratic residue modulo $7$. I only asked the question to see if there some other reasoning. – Rene Schipperus Jun 2 '14 at 20:12
Unfortunately, this answer is (currently) incomplete; it does not suffice to consider only $\pmod{7}$ as it is possible for $x, y, z, t \equiv 0 \pmod{7}$, $x, y, z, t$ not all $0$. One really needs to address why this cannot happen. Perhaps you should update your answer to explicitly address this issue. – Ivan Loh Jun 2 '14 at 20:12

Adding, $7(x^2+y^2)=z^2+t^2$, whence $7 \mid z^2+t^2 \Rightarrow 7 \mid z, t$. Thus $49 \mid z^2+t^2 \Rightarrow 7 \mid x^2+y^2 \Rightarrow 7 \mid x, y$. Then $(\frac{x}{7}, \frac{y}{7}, \frac{z}{7}, \frac{t}{7})$ is a solution as well. We continue in this way to get $7^n \mid x, y, z, t$ for all positive integers $n$, so $x=y=z=t=0$.

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What is your logic going from $7|z^2+t^2$ to $z|z,t$ ? – Rene Schipperus May 31 '14 at 17:33
@ReneSchipperus Either note $-1$ is not a quadratic residue $\pmod{7}$ or enumerate cases. – Ivan Loh May 31 '14 at 17:39
@ReneSchipperus, given $f(x,y) = a x^2 + b x y + c y^2,$ then $\Delta = b^2 - 4ac,$ and a positive odd prime $q$ (not dividing $\Delta$) with $(\Delta|q)= -1,$ if $ a x^2 + b x y + c y^2 \equiv 0 \pmod q,$ then $x,y \equiv 0 \pmod q.$ A bit redundant here, but we usually demand $\Delta$ not a perfect square if non-negative. – Will Jagy May 31 '14 at 18:20
Can you please explain how did you deduced $7^n \mid x, y, z, t$? I Mean I can't get it. – Sagnik Saha Jun 1 '14 at 4:52
@SagnikSaha $7 \mid x, y, z, t$. $(\frac{x}{7}, \frac{y}{7}, \frac{z}{7}, \frac{t}{7})$ also a solution, so the same argument applies and $7 \mid \frac{x}{7}, \frac{y}{7}, \frac{z}{7}, \frac{t}{7}$ so $7^2 \mid x, y, z, t$. Then the same applies to $\frac{x}{7}, \frac{y}{7}, \frac{z}{7}, \frac{t}{7}$ so $7^2 \mid \frac{x}{7}, \frac{y}{7}, \frac{z}{7}, \frac{t}{7}$ so $7^3 \mid x, y, z, t$..... and so on. This can, if need be, be made rigorous via induction. – Ivan Loh Jun 1 '14 at 10:53

For the system of equations:


If the ratio can be represented as: $a=p^2+3$






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