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)

Chatushpathi