Which of the integers cannot be formed with $x^2+y^5$ 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?
 A: I think the answer is 59121. I just used the fact that the last digit of the fifth power of an integer from $0$ to $9$ is the the number itself. Also we have only $5$ digits to chose from the last digits of the square of a number  ,viz, $0,1,4,5,6,9$. Now, for last digit to be 1 I will have to chose from square or fifth power of $10n$ ( for some natural number $n$) And $11$ or vice versa which is not possible for the given range of integers in the options.
Also, 
a) the sum of  fifth power of any of the positive integer with last digit as $6$ and square of any integer with last digit as $5$ 
And 
b) sum of fifth power of any of the positive integer with last digit as $5$ and square of any integer with last digit as $4$  Is not equal to the given number with last digit as $1$ ,ie, $59121$
A: I have found 
$$59170=9^5+11^2$$
$$59012=8^5+162^2$$
$$59149=9^5+10^2$$
$$59130=9^5+9^2$$
and $59121$ can't be expressed in this form.
A: 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).
A: 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.
