## Task $1 / 74$

Determine all natural numbers $n$ for which the expression

$$
T(n)=n^{2}+(n+1)^{2}+(n+2)^{2}+(n+3)^{2}
$$

is divisible by 10 without a remainder.
Obviously, $T(n)$ for any $n$ is always divisible by 2 without a remainder, since exactly two of the four consecutive numbers $n, (n+1), (n+2), (n+3)$ are even and exactly two are odd; the sum of the squares is therefore always an even number. Thus, it suffices to investigate for which natural numbers $n$ the term $T(n)$ is divisible by 5 without a remainder.

If $n \equiv k(\bmod 5)$, then

$$
T(n) \equiv k^{2}+(k+1)^{2}+(k+2)^{2}+(k+3)^{2} \equiv 4 k^{2}+12 k+14 \equiv -k^{2}+2 k-1 \equiv -(k-1)^{2} \quad(\bmod 5)
$$

It follows immediately that $k=1$ must hold if $T(n) \equiv 0(\bmod 5)$ is to be true. Therefore, the natural numbers $n$ are of the form $n=5 m+1$ with $m$ as a natural number, and only these elements belong to the solution set.