2. 

(a) Determine the natural number $n$ for which $401 \cdot 401^{3} \cdot 401^{5} \cdot \ldots \cdot 401^{401}=401^{n^{2}}$.

(b) For $n$ determined in point (a), find the remainder of the division of the number $a=1 \cdot 2 \cdot 3 \cdot \ldots \cdot 100-31$ by $n$.

(c) For $n$ determined in point (a), show that $n^{2015}$ can be written as the sum of three non-zero perfect squares.
Solution.

(a) $401 \cdot 401^{3} \cdot 401^{5} \cdot \ldots \cdot 401^{401}=401^{n^{2}} \Longleftrightarrow 401^{1+3+5+\ldots+401}=401^{n^{2}} \Longleftrightarrow 401^{201^{2}}=401^{n^{2}}, n^{2}=201^{2}$, $n=201$. (3p)

(b) $a=1 \cdot 2 \cdot 3 \cdot \ldots \cdot 100-31=201 \cdot k-201+170=201 \cdot c+170$, so the remainder of the division of $a$ by $n$ is 170. (2p)

(c) $201^{2015}=201 \cdot 201^{2014}=(196+4+1) \cdot\left(201^{1007}\right)^{2}=\left(14 \cdot 201^{1007}\right)^{2}+\left(2 \cdot 201^{1007}\right)^{2}+\left(201^{1007}\right)^{2}$.