Find all prime numbers p such that both numbers $4p^2+1$ and $6p^2+1$ are prime numbers?

If $ pneq 5 $ then $ {rm mod} 5!:, pnotequiv 0,Rightarrow, p,equiv, color{#c00}{pm1}, {rm or} color{#0a0}{pm 2} Rightarrow Bigglbrace begin{align} {p^2equiv color{#c00}1} &Rightarrow,4p^2!+!1equiv 5 equiv 0 \ {{rm or} p^2equiv color{#0a0}4} &Rightarrow,6p^2!+!1equiv 25equiv 0end{align}$

Remark $ $ So it boils down to $bmod 5!: pnotequiv 0,Rightarrow, p^2equiv pm1,,$ a special case of Euler's Criterion.

You might want to double-check your numbers: $4 times 5^2 + 1 = 4 times 25 + 1 = 101$, which is prime, and $6 times 5^2 + 1 = 6 times 25 + 1 = 151$, which is also prime.

But you're right about $2$ and $3$: $4 times 2^2 + 1 = 17$, which is prime, but $6 times 2^2 + 1 = 25$ which is obviously composite; and $4 times 3^2 + 1 = 37$, which is prime, but $6 times 3^2 + 1 = 55$, which is also composite.

Maybe if $p neq 5$ then either $5 mid (4p^2 + 1)$ or $5 mid (6p^2 + 1)$? Well, if $p$ is a prime other than $5$, then $p equiv 2, 3 textrm{ or } 4 pmod 5$. If $p equiv 4 pmod 5$ then $p^2 equiv 1 pmod 5$ and $4p^2 + 1 equiv 0 pmod 5$. But if $p equiv 2 textrm{ or } 3 pmod 5$, then $p^2 equiv 4 pmod 5$ and $6p^2 + 1 equiv 0 pmod 5$.

Ergo, no other prime can satisfy both conditions.

Only $p = 5$ satisfies both $4p^2 + 1$ and $6p^2 + 1$ being prime, giving the primes $101$ and $151$.

For other odd primes either $4p^2 + 1$ or $6p^2 + 1$ can be prime, but not both. Given an odd prime $p neq 5$, we have $p^2 equiv 1$ or $9 bmod 10$. The former gives $4p^2 + 1 equiv 5 bmod 10$ which is clearly composite on account of being divisible by $5$, while the latter gives $6p^2 + 1 equiv 5 bmod 10$, also a multiple of $5$.