Find the minimum $n$ such that $x^2+7=\sum_{k=1}^n f_k(x)^2$ where $f_k(x)\in \mathbb{Q}[x]$ Recently, I have found this problem:

Find the minimum $n \in N$ such that $x^2+7=f_1(x)^2+f_2(x)^2+\cdots+f_n(x)^2$ where $f_1(x),+f_2(x),+\cdots+f_n(x)$ are polynomials with rational coefficients. 

I have tried to solve this problem when $n=2$ using $f_1(x)=(a_1x+b_1)^2$ and $f_2(x)=-(a_2x+b_2)^2$, but I can't go on. Any idea?

Let's assume you can do with only two polynomials. Then we have:
$$
p = \sum_{i=1}^{k_1}{p_nx^n}
\\
q = \sum_{i=1}^{k_2}{q_nx^n}
$$
with
$$
p^2+q^2 =x^2 +7
$$
Therefore, if we can do it with only two polynomials, then there's a solution for
$$x^2+7 = (ax+b)^2 + (cx+d)^2 
\\
x^2+7 = (a^2+c^2)x^2+(2ab+2cd)x + (b^2+d^2)
$$
By equating coefficients and solving the equation system for $a,b,c$, we have 
$$a =\pm \frac d{\sqrt7}$$
And therefore either $a$ or $d$ is irrational.
