utkarshgupta wrote: Do there exist two polynomials with integer coefficients such that each polynomial has a coefficient with an absolute value exceeding $2015$ but all coefficients of their product have absolute values not exceeding $1$? ($10$ points) Let $P_n(x)=\prod_{k=1}^n\left(1-x^{2^k}\right)$ It is easy to check that : 1) all the coefficients of $P_n$ $\in\{-1,0,1\}$ 2) The two highest degree summands of $P_n(x)$ are $(-1)^nx^{2^{n+1}-2}-(-1)^nx^{2^{n+1}-4}$ 3) $Q_n(x)=(1-x)^n|P_n(x)$ and so $P_n(x)=Q_n(x)R_n(x)$ for some $R_n$ 4) The two highest degree summands of $Q_n(x)$ are $(-1)^nx^n-(-1)^nnx^{n-1}$ 5) And so the two highest degree summands of $R_n(x)$ are $x^{2^{n+1}-n-2}+nx^{2^{n+1}-n-3}$ It remains to choose $n$ great enough to prove required existence.

Posted before; see the topic 2015 Moscow Olympiad - Polynomial Product.

Another example: $(x+1)^{2017}$ and $\prod_{i=1}^{2017}{(x^{2d_i}-x^{2d_i-1}+x^{2d_i-2}-...+x^2-x+1)}$ where $d_1,d_2,...,d_{2017}$ are positive integers inductively defined by $d_1=1$ and $2d_k+1>\sum_{i=1}^{k-1}{(2d_i+1)}$ for each $k\in \{ 2,3,....,2017\}$.