and 
be two sequences defined by:![\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]](http://latex.artofproblemsolving.com/a/a/0/aa0a5c5acedc990eb345fb50a8395db05307a4e7.png)
for all 
,
and 
.![\[
a_{2024} < 5.
\]](http://latex.artofproblemsolving.com/f/c/a/fca6e6630f33429b5a7ea867d639a2e29ff881f2.png)
such that the following equation holds:![\[
7^a + 1 = 2b^2
\]](http://latex.artofproblemsolving.com/e/0/8/e08d3fc7feefd7b4acc4a0f9c7e8bcfa328b5791.png)
such that
and 
Prove that
Y by GeoKing, Sarla, SatisfiedMagma
Let 
be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly positive integer factors 
. Assume that no prime larger than 
divides any of the 
's. Show that there must be some perfect cube among the 's.
be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly positive integer factors 
. Assume that no prime larger than 
divides any of the 
's. Show that there must be some perfect cube among the 's.
or 
or 
or 
.
numbers of type 
a
of type 
so there has to be one 
from type 
or 
but any number from there is a perfect cube.
, 
, 
, 
and for second kind there are 
numbers as 
, so still one number lefts and as the numbers are distinct we must have that number to be of other type and rest all of them are perfect cube , hence there is at least one perfect cube among them. 
.
.
.
,
and the other has to be 
which makes it a perfect cube.
,
and 
both of which are divisible by 
,
.
,
.
distinct possible non perfect cube natural possible but we are picking 
of them thus by PHP we will for sure pick one which is a perfect cube.
Let 
denote the number of positive divisors. Using the fact 
we have the following cases,![$$
(e_1(i), e_2(i), e_3(i), e_4(i),e_5(i))=
\begin{cases}
\sigma(6, 16, 16, 0, 0) & \text{$\binom{5}{3}\frac{3!}{2!}=30$ possibilites,}\\[2ex]
\sigma(118, 16, 0, 0, 0) & \text{$\binom{5}{2}2!=20$ possibilities,}\\[2ex]
\sigma(288, 6, 0, 0, 0) & \text{$\binom{5}{2}2!=20$ possibilites,}\\[2ex]
\sigma(2022, 0, 0, 0, 0) &\text{$\binom{5}{1}1!=5$ possibilites.}
\end{cases}
$$](http://latex.artofproblemsolving.com/1/e/7/1e7607b3c2ea2ebe86443b857b4bbd97def78d3c.png)
Where 
just denotes a permutation. Suppose any of the numbers among 
is not a perfect cube then they all fall under the first two cases mentioned above. But there are only a total of 
of them when combined, contradicting our assumption and we are done. 
