If a group is $3$-abelian and $5$-abelian, then it is abelian In a group $(Z,*)$, $(a*b)^{5}=a^{5}*b^{5},\forall a,b\in Z$ and $(a*b)^{3}=a^{3}*b^{3}$ then prove that $Z$ is abelian.
I know that for three consecutive integer if $(a*b)^{i}=a^{i}*b^{i},\forall a,b\in Z$ holds then $Z$ is abelian. 
I know i have to use this property and i have to use three consecutive integer $3,4$ and $5$. But i am stuck.

Here, consider $(G,\circ) \equiv (Z,*)$
Let, $a, b \in G$.
Now, using $a^3 \circ b^3 =(a \circ b)^3$ and $a^5 \circ b^5 =(a \circ b)^5$ :
$(a^3 \circ b^3) \circ (a\circ b)^2 =a^5 \circ b^5$
$ \implies a^3 \circ b^3 \circ (a \circ b)^2 = a^2 \circ (a \circ b)^3 \circ b^2 $ 
$\implies a \circ b^3 \circ (a \circ b)^2 = (a \circ b)^3 \circ b^2 \implies (a\circ b) \circ b^2 \circ (a \circ b)^2 = (a \circ b)^3 \circ b^2 \implies  b^2 \circ (a \circ b)^2 = (a \circ b)^2 \circ b^2$
As $a \circ b= c \in G \forall a, b \in G$, therefore:
$b^2 \circ c^2 = c^2 \circ b^2 \implies c \circ b^2 \circ c^2 \circ b= c^3 \circ b^3 \implies (c\circ b)\circ ( b\circ c) \circ (c\circ b) = (c\circ b)^3 \implies b\circ c = c\circ b$, i.e the Group is abelian
