<s>[INST] Divisors of the form $4n+1$ I read a question
Number of divisors of the form $(4n+1)$
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In the soultion

Any positive divisor of $2^2\cdot 3^3\cdot 5^3\cdot 7^5$ of the form
  $4k+1$ is a number of the form: $$3^a\cdot 5^b\cdot 7^c$$ with $0\leq
> a\leq 3,0\leq b\leq 3,0\leq c\leq 5$ and $\color{red}{a+c}$ being even. There are:
  $$\frac{4\cdot 4\cdot 6}{2}=\color{red}{48}$$

why a+c should be even?
Plz use easy language. I don't know the meaning of mod....
 [/INST] HINT:
As $3=4-1,7=8-1$
$$3^a7^c=(4-1)^a(8-1)^b$$ 
using Binomial Expansion, we can prove it to be of the form $4c+(-1)^{a+b}$ where $c$ is some integer
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