Problem 1186

The answer $n = 3$. In fact, the polynomials $P_{1}(x) = x^{2} - 4$, $P_{2}(x) = x^{2} - 4x + 6$ and $P_{3}(x) = x^{2} - 8x + 12$ satisfy the conditions:

$P_{1} + P_{2} = 2(x - 1)^{2}$,

$P_{1} + P_{3} = 2(x - 2)^{2}$,

$P_{2} + P_{3} = 2(x - 3)^{2}$.

Suppose that there are four polynomials $P_{1}, P_{2}, P_{3}, P_{4}$ satisfying the conditions. Then

$P_{1} + P_{2} = 2(x - t_{12})^{2}$,

$P_{3} + P_{4} = 2(x - t_{34})^{2}$,

$P_{1} + P_{3} = 2(x - t_{13})^{2}$,

$P_{2} + P_{4} = 2(x - t_{24})^{2}$,

where $t_{ij}$ is the multiple root of the polynomial $P_{i} + P_{j}$.

Let $Q = P_{1} + P_{2} + P_{3} + P_{4}$. Then $Q$ has two representations

$$Q = 2(x - t_{12})^{2} + 2(x - t_{34})^{2} \text{ and } Q = 2(x - t_{13})^{2} + 2(x - t_{24})^{2}.$$

By considering linear and constant terms of both expressions of $Q$, we get

$$t_{12} + t_{34} = t_{13} + t_{24} \text{ and } t_{12}^{2} + t_{34}^{2} = t_{13}^{2} + t_{24}^{2}.$$

This shows that $(t_{12}, t_{34}) = (t_{24}, t_{13})$. But, if $t_{12} = t_{13} \Rightarrow P_{2} = P_{3}$, and if $t_{12} = t_{24} \Rightarrow P_{1} = P_{4}$, in both cases, we get a contradiction. $\square$