Can 422215686281216 be expressed as the sum or difference of two fifth powers?

Question

I have a large integer, 422215686281216. I am looking for two fifth powers which when added together, or subtracted from one another, equal this number. How can I tell whether an integer is the sum , or difference, of two fifth powers ?

Answer 1

You can show that sums and differences of two fifth powers cannot be congruent to $3$ modulo $11$.

Then note that your number is congruent to $3$ modulo $11$.

Conclude that you cannot write your number as a sum or difference of two fifth powers.

Added: Oh, so many opinions and helpful comments. To clarify, I merely calculated sums and differences of fifth powers modulo $2, 3, 4,$ etc. until I found a modulus for which not all congruence classes were possible. This happened to be $11$, and it just happened that one class left out, $3$, was the class of the given number. This is a standard method for treating these sums-of-powers questions. For further reading, take a look at most any introductory number theory text (let me know if you would like a more specific reference).

Answer 2

Here's something you could try if Matthew Conroy's trick didn't work.

More generally, suppose you have a positive integer $N$ and you want to write $N = x^5 - y^5$ where $x$ and $y$ are integers (this includes a sum of $5$'th powers if you take $y$ to be negative). Note that $$ N = (x - y)(x^4 + x^3 y + x^2 y^2 + x y^3 + y^4)$$ so $x - y$ must be one of the (positive or negative) divisors of $N$. Enumerate all divisors of $N$ (which you can do from the prime factorization); for each such divisor $d$, check whether the polynomial $(Y+d)^5 - Y^5 - N$ has an integer root. If it does (say $r$), then $y = r$, $x = r+d$ is a solution. If none of these has an integer root, it is not possible to write $N$ as a sum or difference of $5$'th powers.