In my work, I reached the following congruence. Here, $\square$ is a square number. \begin{align} &4+1\equiv\square\text{ modulo }4p, \\&4(p-1)+1\equiv\square\text{ modulo }4p. \end{align} There exists an example each of them hold but I wonder both congruence hold at the same time or not.

For example, if $p=5$, then $5\equiv 5^2$ modulo $20$ but $17\not\equiv\square$ modulo 20.

If $p=3$, then $5\not\equiv\square$ modulo $12$ but $4(3-1)+1\equiv 3^2$ modulo $12$.

I think that both congruence cannot hold at the same time. Can anyone help to prove this?

In my work, I reached the following congruence. Here, $\square_1$, $\square_2$ are square numbers and $p$ denotes the prime number. \begin{align} &4+1\equiv\square_1\text{ modulo }4p, \\&4(p-1)+1\equiv\square_2\text{ modulo }4p. \end{align} There exists an example each of them hold but I wonder both congruence hold at the same time or not.

For example, if $p=5$, then $5\equiv 5^2$ modulo $20$ but $17\not\equiv\square$ modulo $20$.

If $p=3$, then $5\not\equiv\square$ modulo $12$ but $4(3-1)+1\equiv 3^2$ modulo $12$.

I think that both congruence cannot hold at the same time. Can anyone help to prove this?