Martin N. wrote: For a nonnegative integer $n$, define $a_n$ to be the positive integer with decimal representation \[1\underbrace{0\ldots0}_{n}2\underbrace{0\ldots0}_{n}2\underbrace{0\ldots0}_{n}1\mbox{.}\] Prove that $\frac{a_n}{3}$ is always the sum of two positive perfect cubes but never the sum of two perfect squares. (4th Middle European Mathematical Olympiad, Team Competition, Problem 7) I have Found that $\frac{1\underbrace{0\ldots0}_{n}2\underbrace{0\ldots0}_{n}2\underbrace{0\ldots0}_{n}1\mbox{}}{3}=\left(\frac{10^{n+1}-1}{3}\right)^3+\left(\frac{2(10^{n+1}-1)}{3}+1\right)^3$

Martin N. wrote: For a nonnegative integer $n$, define $a_n$ to be the positive integer with decimal representation \[1\underbrace{0\ldots0}_{n}2\underbrace{0\ldots0}_{n}2\underbrace{0\ldots0}_{n}1\mbox{.}\] Prove that $\frac{a_n}{3}$ is always the sum of two positive perfect cubes but never the sum of two perfect squares. (4th Middle European Mathematical Olympiad, Team Competition, Problem 7) $\frac{a_n}{3}=\left(\frac{2\cdot10^{n+1}+1}{3}\right)^3+\left(\frac{10^{n+1}+2}{3}\right)^3$ ,$\frac{a_n}{3}\equiv3(mod4)$

note that $\frac{a_{n}}{3}=(10^{n+1}+1)(\frac{10^{2n+2}+10^{n+1}+1}{3})$ one can find a representation in sum of two cubes with solving these equations: 1. $x+y=10^{n+1}+1$ 2. $x^{2}+y^{2}-xy=\frac{10^{2n+2}+10^{n+1}+1}{3}$

$$\frac{a_n}{3}=\left(\frac{2\cdot10^{n+1}+1}{3}\right)^3+\left(\frac{10^{n+1}+2}{3}\right)^3$$$a_n=3(x^2+y^2)$ $x$ and $y$ different parity.$\implies $ $x^2+y^2 \equiv 1 mod(4)$ $\implies$ $a_n \equiv 3(mod4)$ Contradiction