TITLE: Which of the integers cannot be formed with $x^2+y^5$
QUESTION [13 upvotes]: So, I was asked by my teacher in school to solve this problem it really had me stumped.The problem is as follows:Given that $x$ and $y$ are integers, which of the following cannot be expressed in the form $x^2+y^5$?
$1.)\ 59170$
$2.)\ 59012$
$3.)\ 59121$
$4.)\ 59149$
$5.)\ 59130$
Is it possible for an elegant solution and not tedious trial and error?

REPLY [5 votes]: The use of $y^5$ makes me think that looking at the problem $\bmod 11$ - considering remainders from division by $11$ - would be useful. 
Possible values of $x^2 \bmod 11 $ are $\{0,1,3,4,5,9\}$ and possible values of $y^5 \bmod 11 $ are $\{0,1,10\}\equiv \{-1,0,1\}$
The numbers given $(59170,59012,59121,59149,59130)$ have remainders of $(1,8,7,2,5)$, and the third one of these cannot be reached by the given expression $x^2+y^5$
So: if it is a single-answer question, the answer is certainly option (3).

REPLY [4 votes]: Since $9^5=59049$ the choice $y=9$ can obtain 1.), 4.), 5.) with $x=11,\,10,\,9$. Similarly, $x=162,\,y=8$ addresses 2).
Proving 3) is impossible requires only a modulo 11 analysis; note that $59121=7$. Squares are $0,1,4,9,5,3$. Since 11 is prime, nonzero fifth powers square to 1 by Fermat's Little Theorem, so fifth powers are $0,1,10$. Notice we can't get $7,18$ as sums.