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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 $p=3$?

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    A simple example to illustrate why this cannot work: if $a$ and $b$ are solutions, then so are $-a$ and $-b$. But if $p$ is an odd integer, then this gives different values of $a^p + b^p$.2011-05-17

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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.

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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.

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    For example $1^2+8^2=65$ and $4^2+7^2=65$, but $1^3+8^3=513$ and $4^3+7^3=407$2011-05-17
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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.

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    Not exactly relevant, since the question is about real numbers. The list *$R$eal numbers which are the sum of two squares of real numbers in two or more different ways* is rather uninteresting.2011-05-17
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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 $p$.