How to prove that: $\sqrt{25!+3} \in \mathbb{R}\setminus\mathbb{Q}$

Question

How can I prove that: $$\sqrt{25!+3} \in \mathbb{R}\setminus\mathbb{Q}?$$

Thanks!

Answer 1

The square root of an integer is rational if and only if the integer is a square. $25! + 3$ is not a square since it is $3$ mod $4$.

Answer 2

Suppose there exist two coprime numbers $p,q \in \mathbb{Z} \backslash \{0\}$ such that $$\sqrt{25!+3}= \frac{p}{q},$$

or equivalently $$q^2(25!+3)=p^2.$$

Notice that $25!+3$ is divisible by $3$ but not by $9$. Therefore, $3$ divides $p^2$ that is $3$ divides $p$; we deduce that $9$ divides $q^2(25!+3)$, and finally that $3$ divides $q$. This is a contradiction with $\mathrm{gcd}(p,q)=1$.