Manipulation of conditions of roots of a quadratic equation.

Question

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 $ab$?

Answer 1

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.

Answer 2

$$a^5+b^5=5 a^2 b^2 (a+b)-5 a b (a+b)^3+(a+b)^5$$

Answer 3

$$(a-b)^2=(a+b)^2-4ab$$ 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).

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Let $m:=(a+b)/2,p=ab$, and after simplification,

$$\frac12(a^5+b^5)=m^5+10m^3\left(\sqrt{m^2-p}\right)^2+5m\left(\sqrt{m^2-p}\right)^4=16m^5-5mp^2-20m^3p.$$

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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 $m,p$.

Answer 4

note that $$a^5+b^5=(a+b) \left(a^4-a^3 b+a^2 b^2-a b^3+b^4\right)$$