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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)