MHB Find n in Perfect Square

[Albert1]

$n\in N$ and $n^5+2n^4+2n^3+2n^2+2n+1$ is a perfect square
find $n$

 

[Albert1]

$n\in N$ and $n^5+2n^4+2n^3+2n^2+2n+1$ is a perfect square
find $n$

hint:

Spoiler

$n^4+n^3+n^2+n+1$ is a perfect square , $n=?$
$(n\in N)$

 

[kaliprasad]

My solution

Spoiler

we have $n^5+2n^4+2n^3+2n^2+2n+1= (n+1)(n^4+n^3+n^2+n+1)$
now $n+1$ and $(n^4+n^3+n^2+n+1$ are co-primes as $n^4+n^3+n^2+n+1 = (n+1)(n^3 + n) + 1$
so $(n+1)$ and $(n^4+n^3+n^2+n+1$ have to be perfect squares.
let us check for $(n^4+n^3+n^2+n+1$ find n and then check for n+ 1
We multiply by 16 to make it fraction free
$16(n^4+n^3+n^2+n+1) = (4n^2+2n+1)^2 + (2n+3)^2 + 6> (4n^2+2n+1)^2$
$16(n^4+n^3+n^2+n+1) = (4n^2+2n+2)^2 - 4(n+1)(n-3) < (4n^2+2n+2)^2$ for n not in [-1,3].
and $> 4n^2+2n+2)^2$ for n in $[-1,3]$
checking for n between [1,3] we see that it is a perfect square for n 3. as n is natural number
so n = 3 and n+1 is $2^2$
so only n that satisfies the condition is $n=3$ and for $n=3$ given expression is 484 or $22^2$