fattypiggy123 wrote: Determine the smallest positive integer $n$ such that the following statement is true: There exists $n$ real polynomials, $P_1(x),\ldots,P_n(x)$ such that the sum of any two of them have no real roots but the sum of any three does. Huhhhh ? $n=3$

Oops I meant largest

Can somebody solve this problem?

I claim the answer is $n=3$, with the construction being $P_1(x) = 1, P_2(x)=2, P_3(x)=-3$, which clearly work. Now we show that $n \ge 4$ fails. Suppose otherwise. For each $i$, define $p_i =P_i(0)$. Out of $p_1,p_2, p_3,p_4$, we may WLOG assume that $p_1$ has the largest absolute value. Then $p_1+p_2, p_1+p_3$ and $p_1+p_4$ all have the same sign, again WLOG let it be positive. Suppose any one of $p_2+p_3, p_3+p_4,p_2+p_4$ were positive, say $p_2+p_3$. Since each of $p_1+p_2, p_2+p_3, p_1+p_3$ are positive, and $P_1+P_2, P_2+P_3, P_1+P_3$ have no real roots, we conclude that \[P_1(x)+P_2(x) > 0, P_2(x) +P_3(x) >0,P_3(x) + P_1(x) > 0\]for all $x$. Summing and dividing by 2, we find that $P_1(x) + P_2(x) + P_3 (x) > 0$ for all $x$, hence it has no real roots, contradiction. This implies that $p_2+p_3, p_3+p_4, p_2+p_4$ are all negative, so we can use a similar argument to show that $P_2(x)+P_3(x)+P_4(x)$ has no real roots, contradiction and we are done