Calculating $a^p+b^p$ given $a^2+b^2=c$ If one has $a^2+b^2=c$ where $a$,$b$ and $c$ are real numbers, is there any way to calculate $a^p+b^p$ where $p$ may be any real number? If there is, would you please explain with an example with $p=3$?
 A: There are $a, b, A, B$ such that $a^2+b^2=A^2+B^2$ but $a^p+b^p \ne A^p+B^p$.
So the answer to your first question is NO.
A: The other responses showing you cannot find it just from $a^2+b^2=c$ are correct.  You may be interested in the fact that given $a^2+b^2=c, a+b=d$, you can form $(a^2+b^2)(a+b)=cd=a^3+a^2b+ab^2+b^3$ and  $d^2-c=2ab$, so $cd-\frac{d^3-cd}{2}=a^3+b^3$.  Given as many symmetric polynomials as variables you can find all the higher orders.
A: Since $1^2 + 7^2 = 5^2 + 5^2$, the number $c$ does not uniquely describe $a$ and $b$.  Therefore, you can get not find $a^p + b^p$.
See Numbers which are the sum of two squares in two or more different ways for a list of more such examples.
A: There are times where $c$ does uniquely specify $a,b$, which is maybe worth mentioning.  Suppose $a,b,c$ are all positive integers.  Then we can look at the sum of squares function for $c$, and we know that (up to order) $a,b$ will be unique when $c$ has exactly one prime factor of the form $4k+1$, and each prime factor of the form $4k+3$ appears to an even power.
In other words, if $c$ is of that form, and $a^2+b^2=c$, $a>0$, $b>0$, then $a^p+b^p$ is completely determined for any $p$.   
