[ Higher degree equations (other).]

Vasya added the fourth power and the square of a certain number, different from zero, and told the result to Petya.

Will Petya be able to uniquely determine Vasya's number?
Since the equality $(-a)^{2}+(-a)^{4}=a^{2}+a^{4}$ holds for any $a$, no matter what number Vasya chooses, Petya will not be able to determine it.

## Answer

He will not be able to.