# Sum of polynomials (Tomato subjective 173)

Problem : Let $${{P_1},{P_2},…{P_n}}$$ be polynomials in $${x}$$, each having all integer coefficients, such that $${{P_1}={{P_1}^{2}+{P_2}^{2}+…+{P_n}^{2}}}$$. Assume that $${P_1}$$ is not the zero polynomial. Show that $${{P_1}=1}$$ and $${{P_2}={P_3}=…={P_n}=0}$$

Solution :

As $${P_1},{P_2},…{P_n}$$ are integer coefficient polynomials so gives integer values at integer points.

Now as $${P_1}$$ is not zero polynomial

$${\displaystyle{P_1}(x)>0}$$ for some $$\displaystyle{x \in Z}$$

Then $${\displaystyle{P_1}(x)\ge{1}}$$ or $$P_1(x) \le -1$$ as $${\displaystyle{P_1}(x)}$$=integer

$$\Rightarrow (P_1(x))^2 \ge P_1(x)$$ or $$0 \ge P_1(x) -(P_1(x))^2$$

But it is given that $${{P_1}={{P_1}^{2}+{P_2}^{2}+…+{P_n}^{2}}}$$

This implies $$(P_2 (x))^2+…+(P_n (x) )^2 \le 0$$. This is only possible if  $$(P_1(x))^2 = … = (P_n(x))^2 = 0$$

Hence the values of x for which $$P_1 (x)$$ is non-zero, $$P_2(x) , … , P_n(x)$$ are all zero. The values of x for which $$P_1(x) = 0$$,  we have  $$0=0+(P_2 (x))^{2}+…+(P_n(x))^2$$ implying each is zero.

Therefore $$P_2(x) = … = P_n(x) = 0$$.

Finally $$P_1(x) = (P_1(x))^2$$ implies $$P_1(x) = 0 \text{or} P_1(x) = 1$$. Since $$P_1(x) \neq 0$$ hence it is 1.

(Proved)