Find values of $p$ for which in a field $\mathbb{Z}_p$, two equations have 1 solution.

Question

Find values of $p$ for which in a field $\mathbb{Z}_p$, two equations, say $7x-y=1$ and $11x+7y=3$ have 1 solution.

I can give some values of $p$ like the obvious $p = 7, 11$. But how do I generalize?

What about no solutions??

Answer 1

These linear equations have a single solution if and only if the corresponding matrix $$ A = \pmatrix{7&-1\\11&7} $$ is invertible and this is the case if and only if its determinant $$ \mathrm{det} A = 7 \cdot 7-(-1) \cdot 11 = 60 \in \mathbb{Z}_p $$ is invertible. Therefore you have a single solution if and only if $p$ (is a prime) different from $2$, $3$ and $5$.

Answer 2

If you choose your favourite method of solution, for example $77x-11y=11$ and $77x+49y=21$ so that $60y=10$; then $49x-7y=7$ so that $60x=4$, you only have a problem if dividing by $60$ (the determinant) is equivalent to dividing by zero (i.e. the relevant matrix is singular mod $p$) - so with the primes $2,3,5$ and you simply have to test those to see whether they give multiple or no solutions.

Answer 3

Substituting $y=7x-1$ into $11x+7y=3$ the equations lead to: $$p\mid60x-10=2\times5\times\left(6x-1\right)$$

If $p\in\left\{ 2,5\right\} $ then any $x$ will do, so the number of solutions exceeds $1$.

If $p\notin\left\{ 2,5\right\} $ then $p\mid6x-1$ showing immediately that there are no solutions for $p=3$.

If $p\notin\left\{ 2,3,5\right\} $ and if $x_{1}$ and $x_{2}$ both satisfy, then $p\mid\left(6x_{1}-1\right)-\left(6x_{2}-1\right)=6\left(x_{1}-x_{2}\right)$ leading to $p\mid x_{1}-x_{2}$. This tells us that the number of solutions does not exceed $1$. Secondly $p$ and $6$ are coprime so $6x+pn=1$ for some pair $\left(x,n\right)\in\mathbb{Z}^{2}$, showing that a solution exists.