I need to determine weather the equation $x^5+y^5=47$ has integer solutions. The way i was taught to do this was to take modulo 4 of the entire equation and check if the sides are equal, so taking modulo 4 of the equation I get: $$[x^5]_4+[y^5]_4=[47]_4$$ So the possible values of both $[x^5]$ & $[y^5]$ are $\{0,1,3\}$ and $[47]_4=3$ so there has to be some values of x and y that satisfy that equation right?
Thanks for the help!!!
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There are no solutions. It's easy to see there are no solutions with x and y positive. If one of them is negative, x+y is a proper divisor and 47 is prime, so x and y must differ by 1 in absolute value. This is now also easy to eliminate since adjacent 5th powers never differ by 47.
Also, solvability mod some number is not a sufficient condition for solvability in integers (except in special cases). It is only a necessary condition. So you can only use it to prove a solution in integers does not exist. You can't generally use it to prove a solution exists...you have to do more work for that.
Just picking an arbitrary modulus like $4$ is unlikely to work very well. You want to pick a modulus that will work well with the powers you have.
In this case, a good one to try is $11$, because there are only three possibilities for $x^5 \bmod {11}$: $-1$, $0$, and $1$. (Granted, there are only three possibilities modulo $4$, too, but that's $\frac34$ of all possibilities, whereas here we get only $\frac{3}{11}$ of all possibilities.) In general, we want to pick a modulus $m$ such that $\phi(m)$ is a small multiple of $5$, and it only makes sense to consider primes and powers of primes.
So the possible values of $x^5 + y^5 \bmod{11}$ are $-2,-1,0,1,2$, but $47 \equiv 3 \pmod {11}$, which is not one of the possible values.