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The problem

Find minimum $n$ for $x^2+3x+6=f_1(x)^2+f_2(x)^2+ \cdots +f_n(x)^2$Find the minimum value of $n$ such that there exists $f_i \in \mathbb{Q[i]}, i=1,2,...,n$ such that, \begin{align}x^2+3x+6=f_1(x)^2+f_2(x)^2+ \cdots +f_n(x)^2\end{align}

My attempts What I have been thinking is that $f_i(x)$ must all be of degree $1$. The reason is, assume $f_i(x)= a_kx^k +\cdots$, then $f_i(x)^2$ has the leading coefficient of $a_k^2$, and therefore if $\text{deg}f_i= \max \{\text{deg}f_t \mid t=1,2,\cdots,n$} is not gonna cancel anyway.

So anyway $f_s(x) = (a_sx+b_s)$, but then what? How should I develop from here? Is there any theorem on how rational numbers can/cannot sum up to integers?

Any help is appreciated!

Hakim Blackear
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  • There's always the four square theorem https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem , but I'm not sure that helps for this particular problem. Can you find some $n$ that works? – TomKern Nov 27 '21 at 14:37
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    $4x^2+12x+24=(2x+3)^2+3^2+2^2+1^2+1^2$, so $n=5$ is an upper bound on the minimum $n$. – Barry Cipra Nov 27 '21 at 14:45
  • @Ben Huh? No. Q[i] is rational, not complex sir – Hakim Blackear Nov 27 '21 at 15:27
  • Misread but $\Bbb{Q}[x]$ is the usual way of denoting polynomials over $\Bbb{Q}$ – Ben Nov 27 '21 at 15:30
  • Also using i as an index in 2 different places is a bit confusing – Ben Nov 27 '21 at 15:30
  • Yes - as I said I misread. I made suggestions to make it less likely that someone else will misread – Ben Nov 27 '21 at 15:31
  • @user994373 Yes. 2 things: first I was sorry I did not notice because I already found a solution. second, I did upvote (an appreciation for your efforts), and did not accept the answer because I was looking for something elementary, and therefore not accepting allows me to receive other approaches. – Hakim Blackear Jan 05 '22 at 17:00
  • If you have your own solution you prefer you should post it. – user994373 Jan 06 '22 at 17:20

1 Answers1

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Let $\mathbf{v} = (v_1,\ldots,v_n)$ be a vector with $\|\mathbf{v}\| = 1$ and entries in $\mathbf{Q}$. It is a fact, not completely obvious, that $\mathbf{v}$ extends (as either a row or a column) to an orthogonal matrix with rational entries. (For example, see Is every unit vector in $\mathbb{Q}^n$ the first column of a rational orthogonal matrix?). In particular, there exists an orthogonal matrix $M \in \mathrm{O}(n)$ with $M \mathbf{e}_1 = \mathbf{v}$, or replacing $M$ by its inverse, with $M \mathbf{v} = \mathbf{e}_1$.

The polynomial $x^2 + 3 x + 6$ can clearly be written as the sum of $n$ squares if and only if the same is true for $4(x^2 + 3 x + 6) = (2x + 3)^2 + 15$, and this can clearly be written as the sum of $n$ squares if and only if the same is true for $x^2 + 15$. Now write

$$x^2 + 15 = (v_1 x + u_1)^2 + (v_2 x + u_2)^2 + (v_3 x + u_3)^2 + (v_4 x + u_4)^2.$$

Equating coefficients, we deduce, with $\mathbf{v} = (v_1,v_2,v_3,v_4)$ and $\mathbf{u} = (u_1,u_2,u_3,u_4)$, that

$$1 = v^2_1 + v^2_2 + v^2_3 + v^2_4 = \|\mathbf{v}\|,$$ $$0 = 2 v_1 u_1 + 2 v_2 u_2 + 2 v_3 u_3 + 2 v_4 u_4 = 2 \mathbf{v}.\mathbf{u},$$ $$15 = u^2_1 + u^2_2 + u^2_3 + u^2_4 = \|\mathbf{u}\|.$$

Now take a rational orthogonal matrix $M$ above, so $M \mathbf{v} = \mathbf{e}_1$, and $M \mathbf{u} = \mathbf{x}$ for some rational vector $\mathbf{x}$. Since orthogonal matrices preserve angles and lengths, we get

$$\mathbf{e}_1.\mathbf{x} = 0, \ \|\mathbf{x}\| = 15.$$

The first equality implies that $\mathbf{x} = (0,a,b,c)$ for rational $a$, $b$, and $c$, and the second implies that

$$a^2+b^2+c^2 = 15.$$

This, however, has no solutions in rational numbers because $15 \equiv 7 \bmod 8$ and no such numbers are sums of three rational squares by a standard $2$-adic argument. (When is a rational number a sum of three squares?)

Thus the minimal possible number of squares is $5$, but clearly

$$x^2 + 15 = (x)^2 + 3^2 + 2^2 + 1^2 + 1^2.$$

user994373
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