Is the diophantine equation $a = x^p - y^p$ sufficient to find $x$ and $y$ in terms of $a$ and $p$? Let there be a natural number $a,$ that can be expressed as $a = x^p - y^p$ where $x, y$ and $p$ are natural numbers, each two of them being pairwise co-prime and $p$ is an odd prime. Then can $x$ and $y$ be found out uniquely in terms of $a$ and $p$? If yes, how?
 A: If $x^p-y^p=z^p-w^p$, then $x^p+w^p=y^p+w^p$, so this question is equivalent to asking if there is a number that is the sum of two $p^{th}$ powers in two different ways. It is unknown if there are any examples of this for $n>3$, see: 
Guy, Richard K. (2004). Unsolved Problems in Number Theory (Third ed.). New York, New York, USA: Springer-Science+Business Media, Inc. ISBN 0-387-20860-7.
Thus we restrict our attention to $p=3$. It is a result of Euler that all rational solutions are of the form: $(3a^2+5ab−5b^2)^3+(4a^2−4ab+6b^2)^3+(5a^2−5ab−3b^2)^3 = (6a^2−4ab+4b^2)^3$.
This can be rewritten as $(A^2+7AB−9B^2)^3+(2A^2−4AB+12B^2)^3 = (2A^2+10B^2)^3+(A^2−9AB−B^2)^3$ with $a=A+B, b=A-2B.$ Taking the parts of this equation mod $3$ yields $A^2+AB$, $2A^2-AB$, $2A^2+B^2$, $A^2-B^2$. We want all of these to not be $0$ mod $3$, because otherwise $(3,x)=3$. But it's clear that $B\neq 0$, as otherwise $a=b$, so by checking the other possible values of $A$ and $B$ mod $3$, none of them produce four terms none of which are $0$ mod $3$. Thus the answer is yes, it is unique for $p=3$
