<s>[INST] Number of Divisors of $N = 3^55^77^9$ of the form $4n+1$ I need to find the number of divisors of $N = 3^55^77^9$ that are of the form $4n+1$, $n\geq 0$.
My try:
I noticed that $5$ itself is a number of the form $4n+1$ so all of its power satisfy the required condition, so number for the exponent of $5$ will be $7+1 = 8$.
Also, even powers of $3$ and $7$ satisfy the required condition. 
But, I also noticed that odd powers of $3$ and $7$ multiplied together also satisfy the given condition as $(4m+3)(4k+3) = 4q+1$. How would I take that into account?
 [/INST] A generating function approach:
Let $$f(x)=\left(1+x+\cdots+x^5\right)\left(1+x+\cdots+x^9\right)$$
Then the divisors of $3^57^9$ which are of the form $4n+1$ are the number of $3^a7^c$ with $0\leq a\leq 5, 0\leq c\leq 9$ and $a+c$ is even. This is the sum of the even coefficients of $f(x)$ which can be written as:
$$\frac{1}{2}(f(1)+f(-1))$$
But $f(-1)=0,$ so the result is $\frac{60}{2}=30.$
Then multiply by $7+1$ to get the result for $3^55^77^9.$
In general, if we count the divisors of $3^m7^n$ of the form $4n+1,$ you have that the number is $\frac{1}2(m+1)(n+1)$ if either $m,n$ is odd. When both are even, the you get $f(-1)=1$ and the number of divisors is $\frac{1}{2}\left[(m+1)(n+1)+1\right].$

You can take the set $D$ of all divisors $3^a5^b7^c$ where $a+c$ is even, and you get a map $\phi:D\to D$ defined as $\phi(3^a5^b7^c)=3^{5-a}5^b7^c$ then $\phi(\phi(d))=d$ and $d$ is of the form $4n+1$ iff $\phi(d)$ is not. So that means exactly one half of $D$ are of the form $4n+1.$
That only works because $5$ is odd. 
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