Try this beautiful problem from TOMATO Subjective Problem no. 173 based on the Sum of Polynomials.
Problem : Sum of polynomials
Let [latex] {{P_1},{P_2},...{P_n}}[/latex] be polynomials in [latex] {x}[/latex], each having all integer coefficients, such that [latex] {{P_1}={{P_1}^{2}+{P_2}^{2}+...+{P_n}^{2}}}[/latex]. Assume that [latex] {P_1}[/latex] is not the zero polynomial. Show that [latex] {{P_1}=1}[/latex] and [latex] {{P_2}={P_3}=...={P_n}=0}[/latex]
Solution :
As [latex] {P_1},{P_2},...{P_n}[/latex] are integer coefficient polynomials so gives integer values at integer points.
Now as [latex] {P_1}[/latex] is not zero polynomial
[latex] {\displaystyle{P_1}(x)>0}[/latex] for some [latex] \displaystyle{x \in Z}[/latex]
Then [latex] {\displaystyle{P_1}(x)\ge{1}}[/latex] or [latex] P_1(x) \le -1 [/latex] as [latex] {\displaystyle{P_1}(x)}[/latex]=integer
[latex] \Rightarrow (P_1(x))^2 \ge P_1(x)[/latex] or [latex] 0 \ge P_1(x) -(P_1(x))^2 [/latex]
But it is given that [latex] {{P_1}={{P_1}^{2}+{P_2}^{2}+...+{P_n}^{2}}}[/latex]
This implies [latex] (P_2 (x))^2+...+(P_n (x) )^2 \le 0 [/latex]. This is only possible if [latex] (P_1(x))^2 = ... = (P_n(x))^2 = 0[/latex]
Hence the values of x for which [latex] P_1 (x) [/latex] is non-zero, [latex] P_2(x) , ... , P_n(x) [/latex] are all zero. The values of x for which [latex] P_1(x) = 0 [/latex], we have [latex] 0=0+(P_2 (x))^{2}+...+(P_n(x))^2[/latex] implying each is zero.
Therefore [latex] P_2(x) = ... = P_n(x) = 0 [/latex].
Finally [latex] P_1(x) = (P_1(x))^2 [/latex] implies [latex] P_1(x) = 0 \text{or} P_1(x) = 1 [/latex]. Since [latex] P_1(x) \neq 0 [/latex] hence it is 1.
(Proved)
