* 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*

**Topic:**Theory of Equations**Central idea:**Sum of square quantities zero implies each of the quantities is zero.**Course:**I.S.I. & C.M.I. Entrance Program