Prove that the number a is natural.a = (2^(2n+3)*3^(n+2)*5^(2n+1) + 3^(n+3)*4^(n+2)*5^(2n))/792 n is from N set

We notice that we can find common factors at numerator.
We'll factorize by the numbers that has the lowest
powers.

First, we'll compare the
powers:

3^(n+2) <
3^(n+3)

5^(2n +
1)<5^(2n)

For the beginning, we'll factorize by
3^(n+2)*5^(2n)

a = 3^(n+2)*5^(2n)*[2^(2n + 3)*5 +
3*4^(n+2)]/792

We notice that we can write 4^(n+2) as a
power of 2:

4^(n+2) = 2^2(n + 2) = 2^(2n +
4)

2^(2n + 4) > 2^(2n +
3)

We'll factorize by 2^(2n +
3):

a = 2^(2n + 3)*3^(n+2)*5^(2n)*(5 +
6)/792

a = 2^(2n +
3)*3^(n+2)*5^(2n)*(11)/792

We'll simplify by
11:

a = 2^(2n +
3)*3^(n+2)*5^(2n)/72

We'll re-write the
numerator:

2^(2n + 3) = 2^2n*2^3 =
8*2^2n

3^(n+2) = 3^n*3^2 =
9*3^n

2^(2n + 3)*3^(n+2) =
8*9*2^2n*3^n

2^(2n + 3)*3^(n+2) =
72*2^2n*3^n

a =
72*2^2n*3^n*5^2n/72

We'll simplify by
72:

a = 2^2n*3^n*5^2n

Since n
is natural, then 2^2n is natural, too.

Since n is natural,
then 3^n is natural, too.

Since n is natural, then 5^2n is
natural, too.

The product of 3 natural
numbers is also a natural
number:

a = 2^2n*3^n*5^2n is a
natural number.