# Integer solutions to $z^2=10x^2-2y^2$

Mathematics Asked by Slave of Christ on November 22, 2020

I need to show that a $$(67, 12, 2)$$-design does not exist. I tried to use the Bruck, Ryser and Chowla theorem which in this case states that if a $$(67, 12, 2)$$-design exists, then the equation
$$z^2=10x^2-2y^2,$$
has a non-trivial integer solution $$(x,y,z)neq overline{0}$$. Checking from Wolfram I found that it does not have any, which would prove my original problem, but I would like to see proof why it doesn’t have any.

It is easy to see that $$z$$ should be even so I tried to plug in $$z=:2k$$ and split it in some cases but it did not go anywhere.

## 3 Answers

If $$5$$ divides either $$y$$ or $$z$$, then it divides all three of $$x$$, $$y$$, and $$z$$, in which case it can be factored out, leaving a smaller solution, so we may as well assume that $$5notmid yz$$, in which case $$y^2=10x^2-2z^2$$ implies $$y^2equiv-2z^2$$ mod $$5$$, which can be written as $$3equiv(yz)^2$$ mod $$5$$ (since $$-2equiv3$$ and $$z^4equiv1$$ mod $$5$$). But the only nonzero squares mod $$5$$ are $$1$$ and $$4$$.

Answered by Barry Cipra on November 22, 2020

Note that $$n^2 equiv 0,1,4 (mod 5)$$, which means that $$-2n^2 equiv 0,2,3 (mod 5)$$. Therefore if $$z^2 = 10x^2-2y^2 equiv -2y^2 (mod 5)$$, then $$z$$ and $$y$$ are both divisible by 5, so $$z=5z_1$$ and $$y=5y_1$$ for some integers $$y_1$$ and $$z_1$$. Then equation becomes $$5z_1^2=2x^2-10y_1^2$$, which means that $$x$$ is divisible by 5, so $$x=5x_1$$ for some $$x$$.

Now we get $$z_1^2=10x_1^2-2y_1^2$$ which is of the same form as the original one, so we can repeat these steps infinitely many times and get $$(x,y,z)=(5x_1,5y_1,5z_1)=(5^2x_2,5^2y_2,5^2z_2)=...=(5^nx_n,5^ny_n,5^nz_n)=...$$ . However, the only number which is divisible by any power of 5 is zero, which means that $$(x,y,z)=(0,0,0)$$, so there are no non-trivial solutions.

Answered by Daniyar Aubekerov on November 22, 2020

I believe the following is correct after seeing from Daniyar that one should resubstitute:

By Jyrki's hint, $$z^{2} = 10x^{2} - 2y^{2} equiv -2y^{2} pmod 5$$. It is easily verified that $$y^{2} equiv 0,1,4 pmod 5$$ so that $$-2y^{2} equiv 0,-2,-8 pmod 5equiv 0,2,3 pmod 5$$.

Clearly then as we have $$z^{2}$$ on the $$LHS$$, we have both $$z,y equiv 0 pmod 5 implies 25 | z^{2},y^{2} implies 25| z^{2},2y^{2} implies$$

$$25|z^{2}+2y^{2}implies 25|10x^{2} implies 5|2x^{2} implies 5|x implies x equiv 0 pmod 5$$

Suppose $$a=z,b=y,c =x$$ is the solution to $$z^{2} = 10x^{2} - 2y^{2}$$ where one of $$a,b,c$$ is smallest.

Hence we may find a "smaller" solution by checking with $$a=5k, b=5j$$, and $$c = 5l$$:

As Daniyar has shown me, it is important to substitute this back into the original equation to get $$25k^{2} = 250l^{2} -50j^{2} Longleftrightarrow k^{2} = 10l^{2} -2j^{2}$$ where $$k,j,l < a,b,c$$ respectively, contradicting the assumption that $$(a,b,c)$$ is the smallest solution.

To clarify, this leaves the case where $$a=k, b=j, c=l$$, which from above is true if and only if $$a,b,c=0$$. Since we have found that $$(0,0,0)$$ satisfies $$z^{2} = 10x^{2} - 2y^{2}$$, this is the only solution.

Answered by Derek Luna on November 22, 2020

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