Manipulation of conditions of roots of a quadratic equation. How do I write $a^5+b^5$ in terms of $a+b$ and $ab$. Also is there any general way of writing $a^n+b^n$ in terms of $a+b$ and $ab$?
 A: This is what I found when I first came across this question. Now let me see if I can find the steps to get Lozenges' expression. First of all, Dr. Sonhard's factorization is correct:
$$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4).$$
Were Zlatan's sign doubt correct, we would get, I suppose, either $ab$ or $a-b$ as the first factor, but with $ab$ we get a $a^5b$ term, which is not on the LHS, and with $a-b$ the last term would be $-b^5$, whereas we have $b^5$. In fact, a general way to factoize $a^n+b^n$ with $n$ odd is:
$$a^n+b^n=(a+b)\cdot\sum_{k=0}^{n-1}(-1)^{n-k-1}a^kb^{n-k-1}.$$
Let us verify this:
\begin{align*}
(a+b)\cdot\sum_{k=0}^{n-1}(-1)^{n-k-1}a^kb^{n-k-1}={}&\sum_{k=0}^{n-1}(-1)^{n-k-1}a^{k+1}b^{n-k-1}+\sum_{k=0}^{n-1}(-1)^{n-k-1}a^kb^{n-k}={} \\
{}={}&a^n+\sum_{k=0}^{n-2}(-1)^{n-k-1}a^{k+1}b^{n-k-1}+b^n+\sum_{k=1}^{n-1}(-1)^{n-k-1}a^kb^{n-k}={} \\
{}\underset{\substack{|\\\ell=k+1}}{=}{\hspace{-.35cm}}&a^n+b^n+\sum_{\ell=1}^{n-1}(-1)^{n-\ell}a^\ell b^{n-\ell}+\sum_{k=1}^{n-1}(-1)^{n-k-1}a^kb^{n-k}.
\end{align*}
Notice how the two sums cancel out, since the terms are all identical except for the sign. Note also that if $n$ were even then we would get $-b^n$ instead of $b^n$.
Lozenges says that:
$$a^5+b^5=5 a^2 b^2 (a+b)-5 a b (a+b)^3+(a+b)^5.$$
tatan commented suggesting to use:
\begin{align*}
a^4+b^4={}&(a^2)^2+(b^2)^2 \\
-a^3b-ab^3={}&-ab(a^2+b^2).
\end{align*}
With that, we rewrite:
$$a^5+b^5=(a+b)((a^2)^2+(b^2)^2-ab(a^2+b^2)+a^2b^2).$$
But this does not seem to lead me anywhere, since a sum of squares is something I do not know how to factorize. So let us go back to $a^5+b^5$ and add and subtract $(a+b)^5$, which is pretty natural since we have a sum of fifth powers:
$$a^5+b^5=a^5+b^5-(a+b)^5+(a+b)^5=(a+b)^5-5a^4b-10a^3b^2-10a^2b^3-5ab^4.$$
It seems now pretty natural to collect a $-5ab$:
$$a^5+b^5=(a+b)^5-5ab(a^3+2a^2b+2ab^2+b^3).$$
That looks a lot like a cube there, right? There are 2's instead of 3's, that's all. So we add and subtract $-6ab(a^2b+ab^2)$:
$$a^5+b^5=(a+b)^5-5ab(a^3+2a^2b+2ab^2+b^3)-5ab(a^2b+ab^2)+5ab(a^2b+ab^2)=(a+b)^5-5ab(a^3+3a^2b+3ab^2+b^3)+5ab\cdot ab(a+b).$$
It looked pretty natural to collect an $ab$ in that last term, right? Oh, but this is just Lozenges' expression! So we are done.
Update
Picking up from where I dropped tatan's suggestion:
$$a^5+b^5=(a+b)((a^2)^2+(b^2)^2-ab(a^2+b^2)+a^2b^2).$$
Apply Lozenges' comment to this post:
\begin{align*}
a^5+b^5={}&(a+b)((a^2+b^2)^2-2a^2b^2-ab(a+b)^2+ab\cdot2ab+a^2b^2)={} \\
{}={}&(a+b)((a^2+b^2)^2-ab(a+b)^2+a^2b^2)={} \\
{}={}&(a+b)(((a+b)^2-2ab)^2-ab(a+b)^2+a^2b^2)={} \\
{}={}&(a+b)((a+b)^4-4ab(a+b)^2+4a^2b^2-ab(a+b)^2+a^2b^2)={} \\
{}={}&(a+b)((a+b)^4-5ab(a+b)^2+5a^2b^2)={} \\
{}={}&(a+b)^5-5ab(a+b)^3+5a^2b^2(a+b),
\end{align*}
which is Lozenges' expression, IIRC. So that is another way to get it, and probably what tatan had in mind when writing his comment.
A: $$a^5+b^5=5 a^2 b^2 (a+b)-5 a b (a+b)^3+(a+b)^5$$
A: $$(a-b)^2=(a+b)^2-4ab$$ so that
$$a,b=\frac{a+b\pm\sqrt{(a+b)^2-4ab}}2.$$
Then
$$a^5+b^5=\left(\frac{a+b+\sqrt{(a+b)^2-4ab}}2\right)^5+\left(\frac{a+b-\sqrt{(a+b)^2-4ab}}2\right)^5$$ that you can evaluate by the binomial theorem (every other term will cancel out).

Let $m:=(a+b)/2,p=ab$, and after simplification,
$$\frac12(a^5+b^5)=m^5+10m^3\left(\sqrt{m^2-p}\right)^2+5m\left(\sqrt{m^2-p}\right)^4=16m^5-5mp^2-20m^3p.$$

For the general case,
$$\frac12(a^n+b^n)=\sum_{k=0}^{n/2}\binom n{n-2k}m^{n-2k}\left(\sqrt{m^2-p}\right)^{2k}=\sum_{k=0}^{n/2}\binom n{n-2k}m^{n-2k}\left(m^2-p\right)^{k}.$$
This is a polynomial in $m,p$.
A: note that $$a^5+b^5=(a+b) \left(a^4-a^3 b+a^2 b^2-a b^3+b^4\right)$$
