We need that the discriminant of the quadratic should be a perfect square, i.e., $(a_1+a_2+\cdots+a_n)^4 = x^2 + (a_1^4+a_2^4+\cdots+a_n^4+1)$. Note that $x$ is odd, looking at the parities of all the $a_i$s. Consider the equation modulo $8$. Then the LHS is $0$ or $1$ mod $8$, and RHS is in the range $2$ to $r+2$ where $r$ is the number of odd elements in the set of all $a_i$s. Clearly, we must have $r\ge 6$ and thus $n\ge 6$. For the construction for $n=6$, let the first 4 numbers be 1 and the last two be -1. This concludes the proof.

Proposed by me

how about n = 2. Let a_1 = 2, a_2 = 1. So the equation is x^2 - 18x + 17 = 0, which has roots 1, and 17

bobo123 wrote: how about n = 2. Let a_1 = 2, a_2 = 1. So the equation is x^2 - 18x + 17 = 0, which has roots 1, and 17 $2^4 + 1^4 + 1 = 18$ not $17$

The answer is \(n=6\), achieved by \(a_1=a_2=a_3=a_4=1\) and \(a_5=a_6=-1\). We will prove impossibility for \(n\le5\). If \(\Delta\) denotes the discriminant, then \[\Delta/4=(a_1+\cdots+a_n)^4-\left(a_1^4+\cdots+a_n^4+1\right)\]needs to be a perfect square. Observe that \((a_1+\cdots+a_n)^4\in\{0,1\}\pmod8\) and \(a_1^4+\cdots+a_n^4\in\{0,1,\ldots,n\}\pmod8\), so \(\Delta/4\in\{-n-1,-n,\ldots,0\}\pmod8\). Evidently \(\Delta/4\) is odd, so if it is a perfect square, then \(\Delta/4\equiv1\pmod8\). This is absurd for \(n\le5\).

$(a_1+a_2+a_3+... +a_n)^4\equiv (0, 1)\mod 8$ To have $D=x^2$ here $D$ is discriminant and $x\in N$, $D=(a_1+a_2+a_3+... +a_n)^4-(a_1^4+... +a_n^4+1)\equiv (0, 1,4)\mod 8$ Just observe for $n\leq 5$ We have $(a_1+a_2+a_3+... +a_n)^4\equiv (0, 1)\mod 8$ $(a_1^4+... +a_n^4+1)\leq 6\mod 8$ So in this case $D\not\equiv (0,4,1)\mod 8$ Hence $n\geq 6$

We want discriminant to be a perfect square hence we must have $(a_{1}+a_{2}+\cdots+a_{n})^4-(a_{1}^{4}+a_{2}^{4}+\cdots+a_{n}^{4}+1)=\mathbb{F}^{2}$ for some $\mathbb{F} \in \mathbb{Z}$ also parity of $(a_{1}+a_{2}+\cdots+a_{n})^{2}$ is same as $a_{1}^{4}+a_{2}^{4}+\cdots+a_{n}^{4}$ so we have $\mathbb{F}$ to be odd so $\mathbb{F}^{2} \equiv 1 \pmod 8$ so we have $(a_{1}+a_{2}+\cdots+a_{n})^{4}-(a_{1}^{4}+a_{2}^{4}+\cdots+a_{n}^{4}) \equiv 2 \pmod 8$ $\textbf{\textcolor{red}{Claim:-}}$ $n \leqslant 5$ don't works. $\textbf{\textcolor{blue}{Pf:-}}$ We have $(a_{1}+a_{2}+\cdots+a_{n})^{4} \equiv 0,1 \pmod 8$ If $(a_{1}+a_{2}+\cdots+a_{n})^{4} \equiv 0 \pmod 8$ Then $a_{1}^{4}+a_{2}^{4}+\cdots+a_{n}^{4} \equiv 6 \pmod 8$ , which can't happen $a_{i}^{4} \equiv 0,1 \pmod 8$ for $1 \leqslant i \leqslant 5$ If $(a_{1}+a_{2}+\cdots+a_{n})^{4} \equiv 1 \pmod 8$ Then $a_{1}^{4}+\cdots+a_{n}^{4} \equiv 7 \pmod 8$ but for that is also not possible. So hence claim follows $\square$ Now we show that $n=6$ works , by the motivation of $a_{1}^{4}+a_{2}^{4}+\cdots+a_{n}^{4} \equiv 6 \pmod 8$ for $(a_{1}+a_{2}+\cdots+a_{6})^{4} \equiv 0 \pmod 8$ , so we set $a_{1}=a_{2}=a_{3}=a_{4}=1$ and $a_{5}=a_{6}=-1$ and we get the quadratic as $x^2-8x+7=0$ which has integer roots $1,7$ , hence this construction works , and hence $\boxed{n=6}$ is the minimal such working $n$. $\blacksquare$