Word for word:
The equation $a^4+b^2+c^2=2014$ has a unique integer solution in positive integers. For this solution, find $a+b+c$.
A. 56 B. 58 C. 60 D. 62 E. 64
This is a tricky one. And apparently there's a way to solve it without calculus.
2026-04-17 10:45:01.1776422701
$a^4+b^2+c^2=2014$
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1
Fermat's theorem: A number N is expressible as a sum of $2$ squares if and only if in the prime factorization of N, every prime of the form $(4k+3)$ occurs an even number of times
Knowing this, you start trying $a$ values. $0,1,2$ are not ok, because they give numbers that do not respect Fermat's condition.
$a=3$ gives you $b^2 + c^2 = 2014-81=1933$ that is a prime number, and it's not of the form $4k+3$. Therefore there are two square numbers whose sum is $1933 = 13^2 + 42^2$ (after some trials)
So $13+42+3=58$