How prove this $ab|a^8+b^4+1$ 
show that: there exsit infinite $(a,b)$ such 
  $$ab|a^8+b^4+1$$
  my try: let $a^8+b^4+1=kab,k\in N^{*}$

and I can't work,Thank you
 A: We show that for any solution $(a, b), a, b \in \mathbb{Z}^+$, we can get another solution $(a', b'), a', b' \in \mathbb{Z}^+$ such that $a'+b'>a+b$. This will imply that starting from the solution $(1, 1)$, we can get an infinite sequence of distinct solutions $(a_n, b_n)$, with $2=a_0+b_0<a_1+b_1<a_2+b_2< \ldots$.
Consider $2$ cases.
Case $1$: $a \leq b^2$. Then we have $a \mid b^4+1$ so $b^4+1=ka$ for some $k \in \mathbb{Z}^+$ so $k \mid b^4+1$. Also $b \mid a^8+1 \Rightarrow b \mid k^8(a^8+1)=(ka)^8+k^8=(b^4+1)^8+k^8 \Rightarrow b \mid k^8+1$. Thus $(k, b)$ is another solution, and $k+b=\frac{b^4+1}{a}+b>a+b$ since $a \leq b^2$.
Case $2$: $a>b^2$. Then clearly $a^4>b^8 \geq b$. We have $b \mid a^8+1$ so $a^8+1=lb$ for some $l \in \mathbb{Z}^+$ so $l \mid a^8+1$. Also $a \mid b^4+1 \Rightarrow a \mid l^4(b^4+1)=(bl)^4+l^4=(a^8+1)^4+l^4 \Rightarrow a \mid l^4+1$. Thus $(a, l)$ is another solution, and $a+l=a+\frac{a^8+1}{b}>a+b$ since $b<a^4$.
As such, we get an infinite sequence of distinct solutions $(a_n, b_n)$, with $2=a_0+b_0<a_1+b_1<a_2+b_2< \ldots$, so indeed there are infinitely many positive integer solutions. 
A: we consider 


*

*$a^8$+$a^4$$\equiv$$-$1(mod ab)

*$a^16$+$b^8$+2$a^8$$b^4$$\equiv$1(mod ab)

*SO ab|$a^16$+$b^8$$-$1

*OR ab|$(a^8 -b^4)^2$ $-$1

*FROM THIS ab|($a^8$$-$$b^4$+1)($a^8$$-$$b^4$$-$1)

*so ab|$a^8$-$b^4$+1 or ab|$a^8$$-$$b^4$$-$1 and we got ab|$a^8$+$b^4$+1......9(1)

*FROM THIS WE GET[adding & substrating  respectively from (1)] a|2$b^3$ and b|2$a^7$


from this (a;b)=(2;1) and there is infinitely many solution from a|$b^3$ and b|$a^7$
