Solve the equations in integers:
a) $3 x^{2}+5 y^{2}=345$

b) $1+x+x^{2}+x^{3}=2$.
a) It is clear that $x^{2}$ is divisible by 5 and less than 115. Therefore, it is sufficient to consider the cases $x^{2}=0,25,100$.

b) Factorize the left side: $(1+x)\left(1+x^{2}\right)=2^{y}$. It is clear that both factors on the left side must be powers of two.

If $x$ is even, the left side is odd, hence $y=0$.

If $x$ is odd, the number $1+x^{2}$ gives a remainder of 2 when divided by 4, that is, $1+x^{2}=2$. Therefore, $x=1$.

## Answer

a) $\pm(10, \pm 3)$
b) $(0,0),(1,2)$.