120. Find all integer solutions \(x\) and \(y\) that satisfy the equation \((2 x+5 y+1)\left(2^{|x|}+x^{2}+x+y\right)=105\).
Solution. Since 105 is an odd number, the numbers $2 x+5 y+1$ and $2^{|x|}+x^{2}+x+y$ are also odd. Therefore, $y$ is an even number, and since the number $x^{2}+x$ is even for any $x$, the number $2^{|x|}$ must be odd, which is only possible when $x=0$.

Now we get the equation $(5 y+1)(y+1)=105$, i.e., the number 105 is represented as the product of two factors, the difference of which is divisible by 4. By simple enumeration, we can verify that this is only possible when $y=4$ (105 = 5 * 21).

Thus, the numbers $x=0$ and $y=4$ satisfy the condition of the problem.