Number of Divisors of $N = 3^55^77^9$ of the form $4n+1$

Question

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?

Answer 1

$N = 3^55^77^9$ 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 $\{1,3,5,7,9\}$

Answer 2

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 $\frac{1}{2}\left[(m+1)(n+1)+1\right].$

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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.

Answer 3

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 $240$