2. Solve the equation in natural numbers $n!+3 n+8=k^{2} \quad(n!=1 \times 2 \times 3 \times$ $\ldots \times n)$.
# Solution

For any natural number $n \geq 3 \quad n!+3n+8$ gives a remainder of 2 when divided by 3. A perfect square does not give a remainder of 2 when divided by 3.

Indeed,

$(3m)^{2}=9m^{2}$ - is divisible by 3,

$(3m \pm 1)^{2}=9m^{2} \pm 6m+1$ - gives a remainder of 1 when divided by 3.

Thus, the equation $n!+3n+8=k^{2}$ has no solutions for $n \geq 3$.

For $n=1$ we get $k^{2}=12$ - no natural solutions.

For $n=2$ we get $k^{2}=16$ and $k=4$.

The solution to the equation: $k=4, n=2$

## Criteria

Only the correct answer - 1 point.

The solution contains the correct answer and shows that the left side of the equation gives a remainder of 2 when divided by 3 for any natural number $n \geq 3$ - 2 points.

The correct answer is obtained, the solution process is correct, but there is no proof that a perfect square does not give a remainder of 2 when divided by 3 - 5 points.