RMO Practice Question 1

$n^4 + 2n^3 + 2n^2 + 2n + 1 = (n + 1)^{2} (n^2 + 1)$

now we can make following cases :

$n^2 + 1 = 1$

n = 0

and

$(n + 1)^{2} = 0$

$n = -1$

so no such required n

$\text{The problem is equivalent to the following equation: } \\ n^4 + 2n^3 + 2n^2 + 2n +1 = a^2 \\ \Rightarrow n^4 + 2n^3 + n^2 + n^2 + 2n +1 = a^2 \\ \Rightarrow n^2\cdot \Big(n^2 + 2n + 1\Big) +1\cdot \Big(n^2 + 2n +1\Big) = a^2 \\ \Rightarrow \Big(n^2 + 2n +1\Big)\cdot \Big(n^2 +1\Big) = a^2 \\ \Rightarrow \Big(n + 1\Big)^2\cdot \Big(n^2 +1\Big) = a^2 \\ \Rightarrow \Big(n^2 + 1\Big) = \Big(\frac{a}{n + 1}\Big)^2 \in \mathbb{Z^+} \\ \large{\stackrel{\frac{a}{n+1} = k \in \mathbb{Z^+}}{\Rightarrow}} n^2 + 1^2 = k^2 \\ \downarrow \\ \big(l\cdot2pq\big)^2+\big(l\cdot (p^2-q^2)\big)^2=\big(l\cdot (p^2+q^2)\big)^2 \\ \text{There isn't any non-primary or primary pythagorean triplet} \\ \text{constituted of 1 since: } \\ l\cdot 2pq \neq 1 \text{ and } l\cdot(p^2 - q^2) \neq 1 \text{for l, p, q integers.}$

Same. $( n^2+1) \text{ can be a perfect square if n=0. But 0 is not a natural number. }$

Well divide the equation by n^2 and substitute n+1/n as y. the equation reduces to y^2+2y-1=k^2 can be written as( y+1)^2-2=k^2 as no perfect squares differ in the interval of 2 we say that no required y and hence no n!!!!!

Multiply and divide the whole expression by n^2 and make a quadratic in n+(1/n). Rest you can do yourself. :-)