Show that there is no non-trivial solution for $x^2 + 2z^2 = 10y^2$

Question

I need help with this to show that the result contradicts Bruck-Ryser-Chowla Theorem, which then implies that no biplane of order $10$ exists. At first I tried proving that no non-trivial solution exists solving the equation under modulo of any integer, which then implies that there is no solution in $\mathbb Z$. But I am not getting it. Please could you help? Please show that there is no non-trivial solution for $x^2 + 2z^2 = 10y^2$.

Answer 1

now when a perfect square is divided by 5 , it can only give remainder 1,0,-1.

now for $x^2+2z^2$ be a multiple of 5 (5 divides $10y^2$), it has to that both x and z are multiple of 5(try all other combinations , none will work).

then let x=5k and z=5l $\Rightarrow$ $25k^2+50l^2=10y^2$ $\Rightarrow$ $5k^2++10l^2=2y^2$.(now y has to be a multiple of 5 , let y=5m).

Now it becomes $k^2+2l^2=10m^2$.(meaning we are back to original equation.)

This means we will repeat the process again and again , then x becomes multiple of 5 ,then 25 , then 125......and so on. Hence x cannot be a finite natural number .(same argument can be given for y and z).

Answer 2

Module $5$ the equation becomes $$ x^2+2z^2\equiv0\pmod5. $$ Note that for an integer $n$ not divisible by $5$, we have $n^2\equiv\pm1\pmod5$. Hence $$ x^2+2z^2\equiv\pm1\pm2\not\equiv0\pmod5, $$ for $x,z$ not divisible by $5$.

So any non-trivial solution to the equation must satisfy $5$ divides both $x$ and $z$, and hence $y$. But if $(x, y, z)$ is a solution, then $(x/5, y/5, z/5)$ is another solution, so they are divisible by $5$ again. This implies that $x, y, z$ are infinitely divisible by $5$, which is impossible. So no non-trivial solutions to the equation exist.

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Hope this helps.