Question

If $p>3$is prime, prove that $p^2+2$ is composite.

Short answer

The remainder when $p^2+2$ is divided by 3 = 0 $\left(mod3\right)$. Therefore, $3\left(p^2+1\right)$ and the number is composite. Hence it is proved.

Step-by-step solution

Define function

Consider the given problem. If $p>3$, then $p^2+2$ is composite.

Applying Division Algorithm,

Every number $n$ is of the form$3k,3k+1$ or $3k+2$.$3p$for $p>3$so $p=3k+1$ or $3k+2$.

If $p^2$ is divided by 3, then the remainder for the${\left(3k+1\right)}^2=3\left(3k^2+2k\right)+1$ is 1.

Also, the remainder forrole="math" localid="1646116176806" ${\left(3k+2\right)}^2=3\left(3k^2+4k+1\right)+1$ is 1.

The remainder when$p^2+2$ is divided by 3=0$\left(mod3\right)$ . Therefore,$3\left(p^2+1\right)$ and the number is composite.

Result 

Hence, it is proved that the remainder when $p^2+2$ is divided by 3=0 $\left(mod3\right)$.

Therefore, $3\left(p^2+1\right)$ and the number is composite.