<s>[INST] 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?
 [/INST] The answer of @Mindlack contains a full solution to the problem. This thread just collects some partial results I got while studying the problem, using only the idea from @Mindlack that $7$ is not the sum of $3$ or less squares of rationals.
Partial results
Notice we can only use polynomials with degree at most $1$, since the coeficient of the leading term of $f_i(x)^2$ is always positive.
Easy upper bound. Taking $f_1(x) = x$ and using the idea on the comments that we can write $7 = 1^2 + 1^2 + 1^2 + 2^2$. This gives a way of using $5$ polynomials.
The argument of @Mindlack. Looking only at the constant coeficients modulo $8$, if we used only $3$ polynomials we would have $7d^2 = a^2 + b^2 + c^2$ with $a,b,c,d$ integers and $\mathrm{gcd}(a,b,c,d)=1$. Since squares modulo $8$ are only $0, 1, 4$, this is impossible. This shows that $n \geq 4$. Also, it will be used again on the arguments below.
Using only one linear polynomial.  Using $f_1(x) = ax + b$ implies that $f_1(x)^2 = a^2 x^2 + 2ab x + b^2$ and we need that $a^2 = 1$, $ab=0$, thus $a = \pm 1$ and $b=0$. Since the other $f_i$ are rationals, the argument of @Mindlack implies we need at least $4$ other polynomials in this case.
Using only two linear polynomials. Take $f_1(x) = a_1x + b_1$ and $f_2(x) = a_2x + b_2$, implying that
$$
f_1(x)^2 + f_2(x)^2 = (a_1^2 + a_2^2) x^2 + 2(a_1b_1 + a_2b_2)x + (b_1^2 + b_2^2).
$$
Since the coeficient of $x^2$ needs to be $1$, we should study the solutions of
$$
a_1^2 + a_2^2 = 1 \quad \text{for rational $a_i$}.
$$
Writing $a_i = \tfrac{p_i}{q_i}$ in lowest terms, we have that
$$
p_1^2 q_2^2 + p_2^2 q_1^2 = q_1^2 q_2^2
$$
Notice that in this case we must have $q_1 \mid q_2$, since $p_1$ and $q_1$ do not have common prime factors. Analogously, $q_2 \mid q_1$ and thus they are equal. Now, we are looking at integer solutions for
$$
p_1^2 + p_2^2 = q^2,
$$
that is, Pythagorean triples. Considering
$$
f_1(x) = \frac{p_1}{q} x + b_1 \quad \text{and} \quad f_2(x) = \frac{p_2}{q} x + b_2
$$
we have
$$
f_1(x)^2 + f_2(x)^2 = x^2 + \frac2q (p_1 b_1 + p_2 b_2) + b_1^2 + b_2^2.
$$
Since the term in $x$ should be zero, we have $b_2 = - b_1 \frac{p_1}{p_2}$ and then
$$
f_1(x)^2 + f_2(x)^2 = x^2 + b_1^2 (1 + \frac{p_1^2}{p_2^2}) = x^2 + b_1^2 \frac{q^2}{p_2^2}.
$$
Notice that the independent term is the square of a rational. Then, using the argument of @Mindlack once again, we cannot use only $4$ polynomials. Indeed, this would imply writing $7$ as a sum of $3$ rational squares.
Partial conclusion. If there is an example with only $4$ polynomials, it must used either $3$ or $4$ linear polynomials.
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