Let $a$ be the smallest positive integer such that $g_i(x) = af_i(x) \in \mathbb{Z}[x]$ for every $1\le i\le n$ Then we have $a^2(x^2+7) = g_1(x)^2+g_2(x)^2+...+g_n(x)^2$ Fact 1: A positive integer $n$ can be expressed as the sum of two squares if and only if for every prime $p$ such that $p|n$ and $p\equiv 3\mod 4$ we have that the maximum power of $p$ that divides $n$ is even. Fact 2: A positive integer can be expressed as the sum of three squares if and only if it is not of the form $4^k(8m + 7)$ -- Suppose that $n=1$. Then $f_1(0)^2 = 7$, and $f_1(0)$ is a rational number, which is impossible. -- Suppose that $n=2$. Then $g_1(0)^2 + g_2(0)^2 = 7a^2$. Notice that the maximum power of $7$ that divides $7a^2$ is odd. Then by Fact 1 this is impossible -- Suppose that $n=3$. Then $g_1(0)^2 + g_2(0)^2 + g_2(0)^2 = 7a^2$. Let $a=2^mb$, where $b$ is odd. Then $7a^2 = 4^m(7b^2)$. We have that $b^2\equiv 1 \mod 8$, then $7b^2 \equiv 7 \mod 8$. Then by Fact 2 this is impossible -- Suppose that $n=4$. We have that each $g_i(x)$ must be a polynomial of degree less than or equal $1$. Let $g_i(x) = a_ix+b_i$. Then $(\sum{a_i^2})x^2 + (2\sum{a_ib_i})x + (\sum{b_i^2}) = a^2x^2+7a^2$. Then $\sum{a_i^2} = a^2, \sum{a_ib_i} = 0, \sum{b_i^2} = 7a^2$. Then $\sum{(a_i+b_i)^2} = 8a^2$, and $\sum{(a_i-b_i)^2} = 8a^2$. Since each square is congruent with 0 or 1 $\mod 8$, we have that each of $a_i+b_i$ and $a_i-b_i$ is even. Then each of $a_i,b_i$ is even. This implies that $a$ is even, and that $(a/2)f_i(x) = (1/2)g_i(x)\in\mathbb{Z}[x]$ for every $i$, which contradicts the minimality of $a$. Therefore we have that $n\ge 5$. We have that $x^2+7 = x^2 + 4 + 1 + 1 +1$. Then $5$ is the answer.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=370122

m.candales wrote: we have that each of $a_i+b_i$ and $a_i-b_i$ is even. Then each of $a_i,b_i$ is even Could you explain this deduction better?

Assume $n=4$ is possible, then there exist integers $a,b,c,d,e,f,g,h,m$ (whose gcd is 1) such that \[(ax+b)^2+(cx+d)^2+(ex+f)^2+(gx+h)^2=m^2(x^2+7)\] Take $x=1,3,5$, we get \[(a+b)^2+(c+d)^2+(e+f)^2+(g+h)^2=8m^2\quad...(1)\] \[(3a+b)^2+(3c+d)^2+(3e+f)^2+(3g+h)^2=16m^2\quad...(2)\] \[(5a+b)^2+(5c+d)^2+(5e+f)^2+(5g+h)^2=32m^2\quad...(3)\] Now we use the following lemma: If $8|p^2+q^2+r^2+s^2$, then $p,q,r,s$ are even. The proof is trivial. Using the lemma on (3) twice, we get that $4|5a+b,5c+d,5e+f,5g+h$. Hence $4|a+b,c+d,e+f,g+h$. Combining with (1), we get that $m$ is even. Using the lemma on (2) twice, we get that $4|3a+b,3c+d,3e+f,3g+h$. So $a,c,e,g$ are even, which implies that $a,b,c,d,e,f,g,h,m$ are all even, contradicting the assumption that their gcd is 1.

Thanks Johan Gunardi

This problem appeared as problem 6 on final of polish MO this year. This is my solution, which I figured out during the contest: Suppose that $n=4$. It's easy to see that degree of each $f_i$ is not greater than $1$. Let $f_4 (x) = ax + b$. We can write $f_1 (x)^2 + f_2 (x)^2 + f_3 (x)^2 - 7 = x^2 - (ax+b)^2 = ((a+1)x+b)((1-a)x-b)$. Right-hand side obviously has a rational root $c$, so $7 = f_1 (c)^2 + f_2 (c)^2 + f_3 (c)^2$. But it's well-known that $7$ can't be expressed as a sum of three squares of rational numbers (to prove this, one can consider equation $a^2 + b^2 + c^2 = 7d^2$ modulo 8). We are done (: Do you think it's nice?

To prove $x^2+7$ is not the sum of four squares, one may use quaternions. Since $f_1,f_2,f_3$ and $f_4$ have real coefficients, we have for instance $(i.f_2).(j.f_3)=(i.j).(f_2.f_3)$, so that we are allowed to write $f_1^2+f_2^2+f_3^2+f_4^2=(f_1+i.f_2+j.f_3+k.f_4)(f_1-i.f_2-j.f_3-k.f_4)$ Thus, $x^2+7$ is a product $(Ax+B).(Cx+D)$, where $A,B,C$ and $D$ are "rational" quaternions. Therefore $AC=1$ $AD+BC=0$ $BD=7$ By the first equation, $A$ and $C$ are inverses, which implies $CA=1$. Similarly, $B.\frac{D}{7}=1\implies DB=7$. Multiplying the second equation on the right by $A$, we get $ADA+B=0$. Multiplying again on the right by $D$, we obtain $(AD)^2=-7$. But if $x^2=7$, we have $x.x.\bar x=7.\bar x$, where $\bar x$ denotes quaternions conjugation. Knowing that $x.\bar x\in\mathbb R$, we see that $x$ is a real multiple of $\bar x$. Therefore, $x\in\mathbb R$ or $x$ is a real multiple of $i+j+k$. Both cases easily conduct to a contradiction. @jerzozwierz: yes, nice solution.

My solution involves consideration of cases: 1.This is the trivial case.If $n=1$ then there exists a polynomial $f_1(x)=ax+b \in \mathbb{Q}[x]$ such that $f_1(x)^2=x^2+7$.Then $b=\sqrt{7}$ is irrational.Contradiction. 2.$n=2$:If two polynomials over $\mathbb{Q}[x]$ satisfy our conditions then there exists polynomials $f_1(x)=ax+b,f_2(x)=cx+d \in \mathbb{Z}[x]$ and $g \in \mathbb{Z}-0$ such that $(ax+b)^2+(cx+d)^2=(x^2+7)g^2$.Putting $x=0$ we get $b^2+d^2=7g^2$.If $g$ is even it is easy to see that the highest power of $2$ in $g$ is same as that in $a,c,b,d$,so wlog we may assume $g$ is odd.Then $b^2+d^2=7g^2 \equiv 3\pmod{4}$.Contradiction. 3.$n=3$:Let $f_1(x)=ax+b,f_2(x)=cx+d,f_3(x)=ex+f$ be three polynomials over $\mathbb{Z}[x]$ and $g \in \mathbb{Z}$ such that $(ax+b)^2+(cx+d)^2+(ex+f)^2=(x^2+7)g^2$.Puuting $x=0$ we get $b^2+d^2+f^2=7g^2$.Again we may assume that $g$ is odd,and then $a^2+b^2+c^2 \equiv 7\pmod{8}$.But the sum of three integer squares cannot leave a residue 7 mod 8.Contradiction. 4.$n=4$:This case is a bit different.We assume that there are polynomials $f_1(x)=ax+b,f_2(x)=cx+d,f_3(x)=ex+f,f_4(x)=gx+h$ over $\mathbb{Z}[x]$ and $t \in \mathbb{Z}$ satisfying our condition.Then $(ax+b)^2+(cx+d)^2+(ex+f)^2-7t^2=t^2x^2-(gx+h)^2=(tx-gx-h)(tx+gx+h)$.The right hand side polynomial has a rational root $r$.For $x=r$ we have $(ar+b)^2+(cr+d)^2+(er+f)^2=7t^2$.The remaining argument is similar to case 3. 5.$n=5$:We have $x^2+7=x^2+2^2+1^2+1^2+1^2$ so $n=5$ is our required answer.

Wrong

ISL 2010 N3 The answer is $n=5$, with an equality case $x^2+7=(x)^2+2^2+1^2+1^2+1^2$. Now suppose we have a solution. Note that the lead coefficients of $f_i^2$ are positive, so the degree of the RHS is two times the max degree of the $f_i$s. Therefore, $f_i(x)=a_ix+b_i$ for all $i\in[n]$. Thus, $\sum a_i^2=1$, $\sum a_ib_i=0$, and $\sum b_i^2=7$. So we have rational vectors $a,b\in\mathbb{R}^n$ with $|a|=1$, $|b|=\sqrt{7}$ and $a\cdot b=0$. Extend $a$ to an orthonormal rational basis $a,e_1,e_2,\ldots,e_{n-1}$. We see that $b=p_1e_1+\cdots+p_{n-1}e_{n-1}$ where the $p_i$ are rational, so $|b|^2=p_1^2+\cdots+p_{n-1}^2$. Therefore, it suffices to show that if $p_1^2+\cdots+p_{n-1}^2=7$, then $n\ge 5$. Suppose not, so then there are rational numbers $a,b,c$ (possibly zero) such that $a^2+b^2+c^2=7$. Let $n$ be the lcm of the denominators of $a,b,c$, so we have $p,q,r$ integers with \[p^2+q^2+r^2=7n^2.\]Suppose we are in the case where $|n|$ is the smallest possible. If $n$ is even, then taking mod 4 quickly tells us that $p,q,r$ are even, so dividing all by $2$ gives smaller solution, which is not possible. Otherwise, mod 4 tells us that $p,q,r$ are all odd. So we have $p=2p'1+1,q=2q'+1,r=2r'+1$, so \[p'^2+q'^2+r'^2+p'+q'+r'=\frac{7n^2-3}{4}.\]But $n^2\equiv 1\pmod{8}$ since $n$ is odd, so the RHS is odd. But the LHS is $\sum p'^2+p=\sum \mathrm{even}=\mathrm{even}$, so we have the desired contradiction. Therefore, we must have $n\ge 5$, and we showed its sharp, so we're done.

This solution is due to Michael Ma. The answer is $n=5$, achieved by $x^2 + 2^2 + 1^2 + 1^2 + 1^2$. We now prove that $n = 4$ is not possible. First, note that: Claim: No three rational squares sum to $7$. Proof. Use infinite descent on the Diophantine equation $a^2+b^2+c^2=7d^2$. $\blacksquare$ Then, by choosing any $t \in {\mathbb Q}$ with $f_1(t) = \pm t$ (and such $t$ must exist since $f_1$ is linear), we would get the contradiction $f_2(t)^2 + f_3(t)^2 + f_4(t)^2 = 7$.

The answer is $n = 5$, achieved by the obvious $x^2 + 2^2 + 1^2 + 1^2 + 1^2$. We will begin with a claim: $\textbf{Claim: }$ There are no solutions in integers to the equation $a^2 + b^2 + c^2 = 7d^2$. The proof is by mod $8$. Observe that $x^2 \equiv 0, 1, 4\pmod 8$. Notice that if $d^2 \equiv 1\pmod 8$, then the right hand side is congruent to $7$ and there is no way to sum to $7$. As a result, $d$ is even. From here, we know that $7d^2 \equiv 0, 4\pmod 8$. In either case, it's easy to see that we can't have any $1\pmod 8$ on the left hand side, meaning all of the numbers are even. If we divide all of the $a, b, c, d$ by $2$, then we proceed by infinite descent. $\square$ Let the polynomials be $a_1x + b_1$, etc(max degree is $1$ because duh). Then, we get the following system: \begin{align*} 0 &= a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4 \\ 1 &= a_1^2 + a_2^2 + a_3^2 + a_4^2 \\ 7 &= b_1^2 + b_2^2 + b_3^2 + b_4^2, \end{align*}and the idea is that we can algmanip these to be more useful $\pmod 8$. We get \begin{align*} 8 &= (a_1 + b_1)^2 + (a_2 + b_2)^2 + (a_3 + b_3)^2 + (a_4 + b_4)^2 \\ 8 &= (a_1 - b_1)^2 + (a_2 - b_2)^2 + (a_3 - b_3)^2 + (a_4 - b_4)^2 \\ -6 &= (a_1^2 - b_1^2) + (a_2^2 - b_2^2) + (a_3^2 - b_3^2) + (a_4^2 - b_4^2). \end{align*}Now, sub $x_1 = a_1 + b_1$ and $y_1 = a_1 - b_1$, so these become \begin{align*} 8 &= x_1^2 + x_2^2 + x_3^2 + x_4^2 \\ 8 &= y_1^2 + y_2^2 + y_3^2 + y_4^2 \\ -6 &= x_1y_1 + x_2y_2 + x_3y_3 + x_4y_4. \end{align*}Multiply each equation by integer $d^2$ so that denominators are cleared, so at this point we're basically dealing with integer systems. Taking the first two equations $\pmod 8$, we find that there can be no $1\pmod 8$s, so all $x$ are even. Similarly, all $y$ are even. Now, our third equation is of the form \begin{align*} -6d^2 &= 4 \left( \frac{x_1y_1}{4} + \frac{x_2y_2}{4} + \frac{x_3y_3}{4} + \frac{x_4y_4}{4} \right), \end{align*}but this implies $2 | d$. Now we can substitute $x' = \frac{x}{2}$, $y' = \frac{y}{2}$, and $d' = \frac{d}{2}$. Subbing these in allows us to do infinite descent, and we are done. $\square$.

A different solution without using three squares theorem: Answer is $n=5$. Construction is $$x^2 + 7 = x^2 + 2^2 + 1^2 + 1^2 + 1^2$$Now FTSOC assume $n=4$ suffices. Note all $f_1$ must be linear. Write each $f_i = p_ix + q_i$. Moreover, we clear denominators so that all $p_i$,$q_i$ are integers and they don't have a common factor. So for some $k \in \mathbb Z_{>0}$ we have: \begin{align*} p_1^2 + \cdots + p_4^2 = k^2 \qquad \qquad (1) \\ q_1^2 + \cdots + q_4^2 = 7k^2 \qquad \qquad (2) \\ p_1q_1 + \cdots + p_4q_4 = 0 \qquad \qquad (3) \end{align*}We have two cases according to parity of $k$: $\textbf{CASE 1:} ~$ $k$ is odd. Then some three of $q_i$ are odd and last is $2$ mod $4$. WLOG $q_1 \equiv 2$ (mod $4$). From $(3)$ we get $p_2 + p_3 + p_4$ is even, so $(1)$ gives $p_1$ is odd. Then $(3)$ further gives $p_2 + p_3 + p_4 \equiv 2$ (mod $4$). But $p_2^2 + p_3^2 + p_4^2 \equiv 0$ (mod $8$). This is a contradiction. $\square$ $\textbf{CASE 2:} ~$ $k$ is even. Then we must have $k \equiv 2$ (mod $4$), as otherwise all $p_i,q_i$ would be even. Then $(1),(2)$ give all $p_i,q_i$ are odd. Now assume $p_1 \equiv q_1$ (mod $4$), as otherwise we change all $q_i \to -q_i$. By adding $(1),(2)$ and invoking $(3)$ we get $$(p_1+q_1)^2 + \cdots + (p_4 + q_4)^2 \equiv 0 \pmod{32}$$Note that $(p_i + q_i)^2$ is $4$ (mod $32$) if $4 \mid p_i - q_i$ and $0$ (mod $16$) otherwise. We obtain a contradiction. $\blacksquare$

I show $n = 4$, which is the most basic. Lemma:$8|a^2+b^2+c^2+d^2 \implies 2|a,b,c,d$ Note all $f_i$ must be linear. Write each $f_i(x) = a_ix + b_i$ $$a_i=\frac{x_i}{N},b_i=\frac{y_i}{N}$$$$x_1y_1+x_2y_2+x_3y_3+x_4y_4=0$$$$x_1^2+x_2^2+x_3^2+x_4^2=7N^2$$$$y_1^2+y_2^2+y_3^2+y_4^2=N^2$$$$\sum (x_i+y_i)^2=8N^2$$hence the number $x_i + y_i$ is divisible by 2.$\implies 2|x_i-y_i$ $$\sum(x_i-y_i)(x_i+y_i)=6N^2$$$$4|6N^2\implies 2|N,N=2t$$$$x_i=p_i+q_i,y_i=p_i-q_i$$$$\sum 4(p_i)^2=8N^2$$$$\sum (p_i)^2=2N^2=8t^2$$$$\sum 4(q_i)^2=8N^2$$$$\sum (qp_i)^2=2N^2=8t^2$$So $ p_i $ and $ q_i $ are also evan.$\implies$ $x_i$ and $y_i$ are also evan. $$x_1y_1+x_2y_2+x_3y_3+x_4y_4=0$$$$x_1^2+x_2^2+x_3^2+x_4^2=7N^2$$$$y_1^2+y_2^2+y_3^2+y_4^2=N^2$$these conditions are fulfilled by $\frac{x_i}{2}$ and $\frac{y_i}{2}$ and $\frac{N}{2}$, which means that the process continues indefinitely. contradiction.

We claim that $n=5$ Construction: Set the 5 functions to be $x,2,1,1,1$ Claim: No 3 rational squares can sum to $7$ Proof: Suppose otherwise. Since we are considering rational numbers, we can instead consider proving the altered statement $$a^2+b^2+c^2=7d^2$$ Where $a,b,c,d$ are integers Subclaim:$a,b,c,d$ must all be even Proof: If $d$ even, $a,b,c$ all even or $2$ odd, $1$ even Taking $\mod 4$, if $2$ odd $1$ even, the LHS $\equiv 2 (\mod 4)$ which is absurd If $d$ odd, $a,b,c$ must be $2$ even $1$ odd or all odd. If $2$ even $1$ odd, similarly take $\mod 4$ to force out a contradiction If all odd, we take $\mod 8$. The LHS will be $\equiv 3(\mod 8)$ while the RHS $\equiv 7 (\mod 8)$, contradiction Now using our subclaim, we proceed by infinite descent to get that there are no solutions to $a,b,c$ , which proves our claim Claim: $n \leq 4$ fails Note that all $f_i(x)$ are of the form $a_ix+b_i$ This is because all the leading coefficients when squared are positive, so $$\deg (f_i(x))^2 \leq 2$$$$\deg (f_i(x)) \leq 1$$ Observe that $n=1$ obviously fails $n=3$ Note $$b_1^2+b_2^2+b_3^2=7$$ Using our earlier claim, there are no rational solutions to this equation $n=2$ Observe that this is simply the $n=3$ case but with $b_1=0$, so there are naturally no solutions either $n=4$ WLOG let $f_1(x)=a_1x+b_1$ where $a_1$ is non-zero. Observe that this must exist or else there cannot be an $x^2$ term Now we set $x= \frac{b_1}{1-a_1}$ Observe that this will cause $x^2=f_1(\frac{b_1}{1-a_1})^2$ Therefore, we have $$f_2(\frac{b_1}{1-a_1})^2+f_3(\frac{b_1}{1-a_1})^2+f_4(\frac{b_1}{1-a_1})^2=7$$ Which does not have any solutions as we have previously mentioned

The answer is $n=5$. This is achievable by $f_1(x) = x, f_2(x) = 2, f_3(x) = f_4(x) = f_5(x) = 1$. Now suppose it was possible for $n=4$. Since not all of the polynomials can be constant, WLOG that $f_1(x) = ax + b$, where $a\ne 0$. Setting $x = r = \frac{b}{1-a}$ (or $-\frac{b}{a+1}$ if $a=1$) gives that $f_1(r)^2 = r^2$, so $f_2(r)^2 + f_3(r)^2 + f_4(r)^2 = 7$. Multiplying both sides by common denominators gives that there exist integers $a,b,c,d$ (such that $d\ne 0$) with \[a^2 + b^2 + c^2 = 7d^2\]This is impossible by Legendre's Three Square Theorem. $\blacksquare$

We claim that the answer is 5. This is achieved by $x^2+2^2+1^2+1^2+1^2=x^2+7.$ Note that clearly, each polynomial is linear. Main Claim: 7 is not the sum of three rational squares. Suppose FTSOC that $$7=\frac{a^2}{b^2}+\frac{c^2}{d^2}+\frac{e^2}{f^2}.$$Then, $$(adf)^2+(cbf)^2+(ebd)^2=7(bdf)^2.$$Hence, $7(bdf)^2$ is a sum of three perfect squares. However, 7 times a square is always a power of 4 times something that is 7 mod 8, contradiction. Let $x_0$ be such that $f_1(x_0)=x_0$. Clearly, $x_0$ is rational since all of the functions are linear. Then, $$7=f_2(x_0)^2+f_3(x_0)^2\cdots.$$Since 7 is not the sum of three rational squares, we need at least 5 terms so we are done.

We claim that the answer is $n=5$. Use the following construction: $x^2+2^2+1^2+1^2+1^2$. We now claim that $n=4$ is not possible and by extension $n\le 4$ is not possible. Suppose \[f_1(x)^2+f_2(x)^2+f_3(x)^2+f_4(x)^2=x^2+7\]then pick $f_4(x)=x$ or $f_4(x)=-x$ since one of those has to have a solution then $f_1(x)^2+f_2(x)^2+f_3(x)^2=7$ which means that there must exist positive integers $a,b,c,d$ such that $a^2+b^2+c^2=7d^2$ and such that $\gcd(a,b,c,d)=1$. Clearly, $a^2+b^2+c^2\equiv \{0,1,2,3\}\pmod 8$ and $7d^2\equiv 0\pmod 0,4,7$ so both must be $0$ in this situation. Thus, $a,b,c,d\equiv 0\pmod 8$, a clear contradiction.

@above Wait why must one of the polynomials be $\pm x$?

One must have an intersection with one of those because linear, so simply pick the value of x that makes it work, sorry if unclear. And I made an mistake: the set of possible values for the sum also includes 5 and 6 but that is minor error

The answer is $n=5$, which can be achieved by taking $x^2+2^2+1^2+1^2+1^2$. We show that $n=4$ is not possible. Note that as $f_1$ is linear, there exists $a$ such that $f_1(a)=\pm a$. Substituting $x=a$ yields \[ f_2(a)^2 + f_3(a)^2 + f_4(a)^2 = 7. \]This means that there is a nontrivial solution in integers to \[ x^2+y^2+z^2=7w^2, \]but this is impossible per Legendre's three-square theorem, so there are no solutions for the $n=4$ case.

The answer is $5$, achieved with $x^2 + 2^2 + 3(1^2)$. We show $n = 4$ is impossible, which finishes. Note that there exists $f_k$ with $f_k$ linear (and no quadratics) - set $f_k(x) = ax+b$. Plugging in one of $x = \frac{b}{1-a}$ or $x = \frac{-b}{a+1}$ obtains that there exist rationals $p$, $q$, $r$ such that $p^2 + q^2 + r^2 = 7$, but this is impossible (clear denominators and note that there are no solutions to $a^2+b^2+c^2=7d^2$ by Legendre's three-square theorem). $\square$

The answer is n=5, with the polynomials being x,2,1,1,1 in some order; we'll show that all 1->4 don't work. For $n\in\{1,2,3\}:x=0\implies\exists r_1,r_2,r_3\in\mathbb{R}: r_1^2+r_2^2+r_3^2=7$, which is impossible by Legendre's three square theorem after suitably multiplying by some multiple of their lcm giving a^2+b^2+c^2=7d^2. If n=4, WLOG assume $f_1(x)=ax+b$ with a nonzero; we can do this because if all polynomials have degree <1, you can't get the x^2, and if the largest degree out of any of them is k>1 with coefficient l_k, then l_k^2>0 means that the sum of those polynomials will have $l_k^2x^{2k}$ in the end result (since largest degree won't cancel), with 2k>2, contradiction. Now, it suffices to just choose some x=x_0 s.t. $f_1(x_0)^2=x^2$; in particular, take $\frac{b}{1-a}$ if a is not 1, and if a is one take -b/2. This now reduces to f_2(x_0)^2+f_3(x_0)^2+f_4(x_0)^2=7, which again has no solutions. $\blacksquare$

Amir Hossein wrote: Find the smallest number $n$ such that there exist polynomials $f_1, f_2, \ldots , f_n$ with rational coefficients satisfying \[x^2+7 = f_1\left(x\right)^2 + f_2\left(x\right)^2 + \ldots + f_n\left(x\right)^2.\] Proposed by Mariusz Skałba, Poland İ think my solution is similar to others. Sketch of my solution: $1)$ Answer is $n=5$ which leads up to a trivial solution. Now assume $n<5$ $2)$ Obviously, the degree of $f_i(x)$ is not bigger than $1$. Let $f_i(x)=a_i*x+b_i$ where we let $a_i$ to be $0$ $3)$ We get some congruences that look a bit messy, but clear inservation of $mod 8$ leads to the result. So no solution for $n<5$

short, why N3, solved with NTguy

First note $x^2+2^2+1^2+1^2+1^2=x^2+7$ is a construction for $n=5$, we claim this is optimal. Letting $d$ be the maximal degree over all $n$ functions, it is clear that $f_1\left(x\right)^2 + f_2\left(x\right)^2 + \ldots + f_n\left(x\right)^2$ has degree $2d$. Thus every polynomial is linear or constant. If $n\leq 4$ then we have $x^2+7=(ax+b)^2+(cx+d)^2+(ex+f)^2+(gx+h)^2$ for some rational coefficients $a$ through $h$. But then observe $a^2+c^2+e^2+g^2=1$ and $ab+cd+ef+gh=0$, thus we have $(ax+b)^2+(cx+d)^2+(ex+f)^2+(gx+h)^2=x^2+(eb+gd-af-ch)^2+(cb-ad-gf+eh)^2+(gb-ed+cf-ah)^2$, and thus $7=w^2+y^2+z^2$ for rational numbers $w$, $y$, $z$, contradicting with Legendre's three-square theorem.