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.

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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