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?
 A: 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. 
A: $N = 3^55^77^9$  gives $(7+1)((3)(5)+(3)(5)) =240$ the first $(3)(5)$ deals with $\{0,2,4\}$ and $\{0,2,4,6,8\}$ the second with $\{1,3,5\}$and $\{1,3,5,7,9\}$
A: Any divisor of $N$ will be of the form $F(x_1,x_2,x_3)=3^{x_1}5^{x_2}7^{x_3}$ where $x_i \in I$
And $0\le x_1 \le 5$ ; $0\le x_2 \le 7 $ ; $ 0\le x_3 \le 9$
$$. $$
Number of divisor are Number of combination of $x_1 ; x_2;x_3$ that are possible such that $F$ is of form $4n+1$.
$$F=3^{x_1}5^{x_2}7^{x_3} =(4-1)^{x_1}(4+1)^{x_2}(8-1)^{x_3}$$
$$F=(4n_1+(-1)^{x_1}).(4n_2+1).(8n_3+(-1)^{x_3})$$
Hence $n(x_2)=8$ and
$$F=4n+(-1)^{x_1+x_3}$$
Hence $x_1+x_3 $ Should be even integer including zero.
Hence Number of divisor are 
$$Case (I) \,\, when \,\, x_1;x_3 \, both \, odd\, n(x_1).n_(x_2).n(x_3)=3×8×5=120 $$
$$Case (II). When \, both \, x_1,x_3 \,are\,even \, n(x_1).n(x_2).n(x_3)=3×8×5=120$$
Hence total number of divisors of the form $4n+1$ are $240$
