# How did Euler realize $x^4-4x^3+2x^2+4x+4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$?

How did Euler find this factorization?

$$\small x^4 − 4x^3 + 2x^2 + 4x + 4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$$

where $\alpha = \sqrt{4+2\sqrt{7}}$

I know that he had some super powers, like he was sent to us from a super intelligent alien universe just to humiliate our intelligence, but how the hell did he do that three centuries ago? :|

• – math.n00b May 6 '14 at 13:29
• How did Ramanujan know $$\frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}$$? – evil999man May 6 '14 at 13:30
• Maybe he solved the quartic and then factorized it! – evil999man May 6 '14 at 13:35
• I deleted my answer as I had made a calculation error... Now I am certain that he had some connections with the Satan. – evil999man May 6 '14 at 14:04
• Perhaps he depressed the polynomial by making the substitution $x=\alpha+1$ and then proceeded from there as suggested by the quartic formula? – illysial May 6 '14 at 14:30

Euler lived a century after Isaac Newton and Blaise Pascal, so he must have been familiar with the former's binomial theorem and the latter's triangle. Indeed, the polynomial you presented looks quite similar to the binomial expansion of $$(x-1)^4$$, whose coefficients are found on the fourth row of Pascal's triangle. By subtracting the two, we are left with $$4x^2-8x-3$$, whose roots are $$1\pm\dfrac{\sqrt7}2$$ which is a quarter of $$\alpha$$. So, $$P(x)=(x-1)^4-4\bigg[(x-1)-\dfrac{\sqrt7}2\bigg]\bigg[(x-1)+\dfrac{\sqrt7}2\bigg],$$ which, after substituting $$u=(x-1)^2$$, becomes $$u^2-4u+7$$. Then, by completing the square, we arrive at the desired result.

• WOW. This is a beatiful answer! I wish I could upvote it a 1000 times. – math.n00b May 6 '14 at 14:45
• You are confusing something. The binomial theorem, in the form of its low degree instances, goes back to the 15th century, if not via Leonardo da Pisa, called "son of the benefactor", in the 13th century to arabic and roman-greek sources even earlier. The binomial series for rational exponents was introduced by Newton and is named sometimes after him. As a special case he certainly observed that this series reproduces the integer binomial formulas. – Lutz Lehmann May 9 '14 at 10:21
• Very Nice answer +1. – Bumblebee Jul 27 '15 at 10:02
• So did people really not know that $(a+b)^n = \sum ...$ before Newton? – Ovi Dec 16 '18 at 1:26
• @Ovi Yup. Omar Khayyam, the Persian mathematician and astronomer of the 11th century knew the theorem before Newton. He even had drawn what is known as Pascal's triangle in his notes. It is speculated that he was also aware of the fact that $n$ doesn't have to be an integer because he could approximate square and cubic roots. – stressed out Nov 15 '19 at 20:36

I like this rather old question. Here is a yet another possible way Euler could have taken:

Note that $\displaystyle x^4+ax^2+b$ can be factorized easily if $\displaystyle a^2-4b\geq 0$. If, however, $\displaystyle a^2-4b\leq 0$, then \begin{align} x^4+ax^2+b&=(x^2+\sqrt{b})^2-(x\sqrt{2\sqrt{b}-a})^2\\ &=(x^2+\sqrt{b}-x\sqrt{2\sqrt{b}-a})(x^2+\sqrt{b}+x\sqrt{2\sqrt{b}-a}). \end{align}

Now in $P(x)=x^4-4x^3+2x^2+4x+4$, use $x=y+1$, and proceed: \begin{align} P(y+1)&=y^4-4y^2+7\\ &=(y^2+\sqrt{7}-y\sqrt{2\sqrt{7}+4})(y^2+\sqrt{7}+y\sqrt{2\sqrt{7}+4}). \end{align}

substituting $y=x-1$ we arrive at Euler's result.

• +1 since you were the closest to achieving Euler's method from scratch. I cheated. I just looked it up. – dustin Feb 16 '15 at 19:37
• @dustin many thanks for the link to the book. that should be fun to read! – Math-fun Feb 16 '15 at 19:56
• The really great thing is Google made the chapters hyperlinkable if you haven't taken a look at it yet – dustin Feb 16 '15 at 19:58
• @dustin I also just found a pdf online! take a look here: archive.org/details/elementsofalgebr00eule – Math-fun Feb 16 '15 at 20:04
• It is the same text. You would have to scroll to chapters since it is a scanned copy without hyperlinking. It is nicer since it has color though. – dustin Feb 16 '15 at 20:07

Once it is noticed that $$x^4-4x^3+2x^2+4x+4=(x-1)^4-4(x-1)^2+7$$ the square can be completed to get $$\left((x-1)^2-2\right)^2+3=\color{#00A000}{\left((x-1)^2-2-i\sqrt3\right)}\color{#0000FF}{\left((x-1)^2-2+i\sqrt3\right)}$$ Solving for $(\alpha+i\beta)^2=2+i\sqrt3$, we get $\alpha^2-\beta^2=2$ and $2\alpha\beta=\sqrt3$. Adding the squares and taking the square root gives $\alpha^2+\beta^2=\sqrt7$. Solving for $\alpha$ and $\beta$ yields $$\color{#00A000}{\left({\small\sqrt{\frac{2+\sqrt7}2}}+i\,{\small\sqrt{\frac{-2+\sqrt7}2}}\right)^2=2+i\sqrt3}$$ $$\color{#0000FF}{\left({\small\sqrt{\frac{2+\sqrt7}2}}-i\,{\small\sqrt{\frac{-2+\sqrt7}2}}\right)^2=2-i\sqrt3}$$ The full factorization is $$\overset{\underbrace{\color{#0000FF}{\left[x-1-\sqrt{\frac{2+\sqrt7}2}+i\,\sqrt{\frac{-2+\sqrt7}2}\right]}\color{#00A000}{\left[x-1-\sqrt{\frac{2+\sqrt7}2}-i\,\sqrt{\frac{-2+\sqrt7}2}\right]}}_{}} {\left[\left(x-1-\sqrt{\frac{2+\sqrt7}2}\right)^2+\frac{-2+\sqrt7}2\right]} \overset{\underbrace{\color{#00A000}{\left[x-1+\sqrt{\frac{2+\sqrt7}2}+i\,\sqrt{\frac{-2+\sqrt7}2}\right]}\color{#0000FF}{\left[x-1+\sqrt{\frac{2+\sqrt7}2}-i\,\sqrt{\frac{-2+\sqrt7}2}\right]}}_{}} {\left[\left(x-1+\sqrt{\frac{2+\sqrt7}2}\right)^2+\frac{-2+\sqrt7}2\right]}$$ This gives the factorization sought.

The other answers are great but were pure speculation since Euler published how he solves quartics.

From Elements of Algebra by Euler section 4 chapter 15, his new method for resolving fourth order equations. Additionally, Google books has a full copy to download or read online here.

Suppose the root of the equation is of the form $x = \sqrt{p} + \sqrt{q} + \sqrt{r}$ where $p,q,r$ are the roots of an equation of degree three, $$z^3 -fz^2+gz-h=0$$ and \begin{align} f&=p+q+r\tag{1}\\ g&=pq+pr+qr\tag{2}\\ h&=pqr\tag{3} \end{align} Now, he square $x$ and obtained $$x^2 = p+q+r+2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}$$ From $(1)$, we can now write $$x^2-f=2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}\tag{4}$$ Let's square $(4)$ again to obtain: $$x^4-2fx^2+f^2 = 4pq+4pr+4qr+8\sqrt{p^2qr}+8\sqrt{pq^2r}+8\sqrt{pqr^2}$$ From $(2)$, we can now write $$x^4-2fx^2+f^2 -4g= 8\sqrt{pqr}(\sqrt{p}+\sqrt{q}+\sqrt{r})$$ Using the identity first laid out for $x$ and $(3)$, we have $$x^4-2fx^2 -8x\sqrt{h}+f^2-4g= 0\tag{5}$$ Euler says we don't need to worry about $yx^3$ because

for we shall afterwards shew, that every complete equation may be transformed into another, from which the second term is taken away.

Next, Euler goes onto factoring a quartic as a product of two real quadratics.

It is to be observed that the product of these three terms, or $\sqrt{pqr}$, must be equal to $\sqrt{h}=b/8$, and that if $b/8$ is positive, the product of the terms $\sqrt{p},\sqrt{q},\sqrt{r}$; must likewise be positive so that all variations that can be admitted are reduced to the four following:

\begin{align} x &= \sqrt{p} + \sqrt{q} + \sqrt{r}\\ x &= \sqrt{p} - \sqrt{q} - \sqrt{r}\\ x &= -\sqrt{p} + \sqrt{q} - \sqrt{r}\\ x &= -\sqrt{p} - \sqrt{q} + \sqrt{r} \end{align} When $b/8$ is negative, we have \begin{align} x &= \sqrt{p} + \sqrt{q} - \sqrt{r}\\ x &= \sqrt{p} - \sqrt{q} + \sqrt{r}\\ x &= -\sqrt{p} + \sqrt{q} + \sqrt{r}\\ x &= -\sqrt{p} - \sqrt{q} - \sqrt{r} \end{align}

In order to use Euler's method, you would first need to transform your equation such that the $x^3$ term is gone. Then you match up the coefficients of $f,g,h$ in your new polynomial with the general form $(5)$. That is, given a polynomial of the form $$x^4 - ax^2 - bx - c = 0$$ you would set $a = 2f$, $b = 8\sqrt{h}$, and $-c = f^2 - 4g$. Then you can follow his book's examples which you will find in the second hyperlink.

• Any reason for the down vote? – dustin Feb 16 '15 at 20:34
• I am not sure why the site let people downvote answers without a comment! This practice discourages people from answering sometimes and leaves others frustrated when they don't know what is wrong with their answer. – NoChance Jul 5 '19 at 1:56

Hint:As @illysial, @Mehdi and others indicated, the substitution $x = t+1$ transforms the expression $x^4 − 4x^3 + 2x^2 + 4x + 4$ into $t^4-4 t^2+7$.

• This is a nice one which I like ;-) This answer was given below. – Math-fun Feb 16 '15 at 19:17
• @Mehdi: Totally, oops. Indeed it was :-) – orangeskid Feb 16 '15 at 19:19