Finding Perfect Squares with $n^4 + 33$

[magneto1]

Find, with proof, all the positive integers $n$ such that $n^4 + 33$ is a perfect square.

 

[kaliprasad]

Find, with proof, all the positive integers $n$ such that $n^4 + 33$ is a perfect square.

Spoiler

let $x^2 = n^4+ 33$

so $33 = (x^2-n^4) = (x+n^2)(x-n^2)$ = $33*1$= $11*3$

gives 2 solutions to be checked

$x+n^2= 33$ and $x-n^2= 1$ => $x= 17$ and $n^2 = 16$ which is a perfect square so n = 4

or

$x+n^2= 11$ and $x-n^2= 3$ => $x= 7$ and $n^2 = 4$ which is a perfect square so n = 2

so we get $2^4 + 33 = 49= 7^2$

and $4^4 + 33 = 289= 17^2$

n = 2 or 4

 

[magneto1]

Nicely done.

 

[Nono713]

Spoiler

Alternatively, $n^4 = (n^2)^2$ and so eventually $(n^2 + 1)^2 - (n^2)^2 > 33$ at which point $n^4 + 33$ is strictly between two perfect squares (and so cannot be one). This upper bound is attained when:
$$(n^2 + 1)^2 - (n^2)^2 > 33 ~ ~ ~ \iff ~ ~ ~ 2n^2 + 1 > 33 ~ ~ ~ \iff ~ ~ ~ n > 4$$
Hence it suffices to check $0 \leq n \leq 4$, and we quickly find that:
$$n = 0 ~ ~ ~ \implies ~ ~ ~ 0^4 + 33 = 33$$
$$n = 1 ~ ~ ~ \implies ~ ~ ~ 1^4 + 33 = 34$$
$$n = 2 ~ ~ ~ \implies ~ ~ ~ 2^4 + 33 = 49 = 7^2$$
$$n = 3 ~ ~ ~ \implies ~ ~ ~ 3^4 + 33 = 114$$
$$n = 4 ~ ~ ~ \implies ~ ~ ~ 4^4 + 33 = 289 = 17^2$$
So the solutions are $n = 2$, $n = 4$.

 

[kaliprasad]

Spoiler

Alternatively, $n^4 = (n^2)^2$ and so eventually $(n^2 + 1)^2 - (n^2)^2 > 33$ at which point $n^4 + 33$ is strictly between two perfect squares (and so cannot be one). This upper bound is attained when:
$$(n^2 + 1)^2 - (n^2)^2 > 33 ~ ~ ~ \iff ~ ~ ~ 2n^2 + 1 > 33 ~ ~ ~ \iff ~ ~ ~ n > 4$$
Hence it suffices to check $0 \leq n \leq 4$, and we quickly find that:
$$n = 0 ~ ~ ~ \implies ~ ~ ~ 0^4 + 33 = 33$$
$$n = 1 ~ ~ ~ \implies ~ ~ ~ 1^4 + 33 = 34$$
$$n = 2 ~ ~ ~ \implies ~ ~ ~ 2^4 + 33 = 49 = 7^2$$
$$n = 3 ~ ~ ~ \implies ~ ~ ~ 3^4 + 33 = 114$$
$$n = 4 ~ ~ ~ \implies ~ ~ ~ 4^4 + 33 = 289 = 17^2$$
So the solutions are $n = 2$, $n = 4$.

Spoiler

as 33 is 1 mod 4, n has to be even because $n^4+33$ cannot be 2 mod 4, so we need not check for odd n so check for 0,2,4 only