Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that $$p(x,y) = x^4 + y^4 + x^2 + y^2$$ can be written as a sum of squares of three polynomials over $x,y$ for real numbers.

share|cite|improve this question
I thought it would be easy.I posed this question to my teacher as well.He was unable to solve it.I was not able to arrive at the answer – Sai Krishna Deep Dec 1 '12 at 7:48
Actually, I can think of a pretty good reason that some of the coefficients might need to be irrational, square roots of positive whole numbers or possibly fractions. I'm not sure yet, though, I will fiddle with it tomorrow. It is late here. – Will Jagy Dec 1 '12 at 7:49
Its from a standardized book in India called "Challenge and Thrill of Pre-College mathematics" – Sai Krishna Deep Dec 1 '12 at 7:55
I'm no mathematician, but I can't see how real numbers can suffice. If I take a linear combination of $x^2, y^2, x, y, xy, 1$ and square these, then I get a positive-definite expression for the $x^2y^2$ and $xy$ terms. Therefore it cannot disappear in the reals?! – Gerenuk Dec 1 '12 at 9:35
up vote 10 down vote accepted

$(\sqrt{2\sqrt{2}-2}y^2+x)^2 + (-\sqrt{2\sqrt{2}-2}xy + y)^2 + (x^2 + (1-\sqrt{2})y^2)^2 = x^4+y^4+x^2+y^2$.

In fact, there are many other solutions by using brute force, as suggested by Will Jagy in the comments.

Step 1 Let's suppose that $x^4+y^4+x^2+y^2 = \sum_{i=1}^3 (a_i x^2 + b_i xy + c_i y^2 + d_i x + e_i y)^2$. If we write $\vec{a} = (a_1,a_2,a_3)$ and so forth, we obtain the following identities by comparing coefficients:

Part (a): $x^2: \| \vec{d} \|^2 = 1$, $y^2: \| \vec{e} \|^2 = 1$, $xy: \langle \vec{d}, \vec{e} \rangle = 0$.

Part (b): $x^3: \langle \vec{a}, \vec{d} \rangle = 0$, $y^3: \langle \vec{c}, \vec{e} \rangle = 0$, $x^2y: \langle \vec{b}, \vec{d} \rangle = - \langle \vec{a}, \vec{e} \rangle$, $xy^2: \langle \vec{b} ,\vec{e} \rangle = - \langle \vec{c}, \vec{d} \rangle$.

Part (c): $x^4: \| \vec{a} \|^2 = 1$, $y^4: \| \vec{c} \|^2 = 1$, $x^2y^2: \| \vec{b} \|^2 + 2 \langle \vec{a}, \vec{c} \rangle = 0$, $x^3y: \langle \vec{b}, \vec{a} \rangle = 0$, $xy^3: \langle \vec{b}, \vec{c} \rangle = 0$.

(All inner products are standard Euclidean inner products)

Step 2 We consider an orthonormal basis of $\mathbb{R}^3$: $\vec{d}, \vec{e}, \vec{f}$. This used part (a).

Using part (b), we get

$\vec{a} = x_1 \vec{e} + y_1 \vec{f}$,

$\vec{b} = -x_1 \vec{d} - x_2 \vec{e} + y_3 \vec{f}$,

$\vec{c} = x_2 \vec{d} + y_2 \vec{f}$

for some reals $x_1,x_2,y_1,y_2,y_3$.

Step 3 Part (c) tells us that

$x_1^2 + y_1^2 = 1$, $x_2^2 + y_2^2 = 1$. (From norm of $\vec{a}, \vec{c}$ being 1.)

$y_1y_3 = x_1x_2$, $y_2y_3 = x_1x_2$ (From $\vec{b}$ perpendicular to $\vec{a}, \vec{c}$.)

$x_1^2+x_2^2+y_3^2 + 2y_1y_2 = 0$.

In the second equation, if $y_3$ is not 0, then $y_1 = y_2$, which forces $x_1 = x_2 = y_1 = y_2 = y_3 = 0$ in the third equation, contradicting the first equation.

So $y_3 = 0$, which implies $x_1x_2 = 0$. WLOG assume $x_1 = 0$, then $y_1 = \pm 1$. The third equation becomes

$0 = x_2^2 + 2y_1y_2 = 1 - y_2^2 + 2y_1y_2 = 2 - (y_2 - y_1)^2$.

So $y_2 - y_1 = \pm \sqrt{2}$. The first equation tells us that $y_1,y_2$ has norms less than 1, which implies that $y_2 = \pm (1 - \sqrt{2})$. This gives $x_2 = \pm \sqrt{2\sqrt{2} - 2}$.

To summarize, up to symmetry of $(x_1,y_1) \leftrightarrow (x_2,y_2)$, we have $x_1 = 0$, $y_1 = \pm 1$, $y_2 = \pm (1 - \sqrt{2})$ (same sign as $y_1$), $x_2 = \pm \sqrt{2\sqrt{2} - 2}$ and $y_3 = 0$.

Step 4 By picking arbitrary orthonormal basis $\vec{d}, \vec{e}, \vec{f}$, we can substitute the values for $x_1,x_2,y_1,y_2,y_3$ above to get solutions for $\vec{a}, \vec{b}, \vec{c}$. All these would give the required sum of squares decomposition. The decomposition I took above comes from choosing the standard orthonormal basis of $\mathbb{R}^3$.

share|cite|improve this answer
Works for me.${}{}$ – Will Jagy Dec 1 '12 at 10:21
I'm not in a position to fully follow this.But thanks for the solution. – Sai Krishna Deep Dec 2 '12 at 6:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.