Divisors of a Number: n2+1 for 4k+1 Divisibility

[jeedoubts]

1. Homework Statement
let N= (7^n1)(9^n2)(11^n1)
if n1 is even,then the total number of divisors of the form of 4k+1 must be divisible by?

correct answer is n2+1

3. The Attempt at a Solution
total number of divisors = (n1+1)(n2+1)(n1+1)

 

[jeedoubts]

1. Homework Statement
let N= (7^n1)(9^n2)(11^n1)
if n1 is even,then the total number of divisors of the form of 4k+1 must be divisible by?

correct answer is n2+1

3. The Attempt at a Solution
total number of divisors = (n1+1)(n2+1)(n1+1)

can anybody please help me uot in this question?

 

[jeedoubts]

1. Homework Statement
let N= (7^n1)(9^n2)(11^n1)
if n1 is even,then the total number of divisors of the form of 4k+1 must be divisible by?

correct answer is n2+1

3. The Attempt at a Solution
total number of divisors = (n1+1)(n2+1)(n1+1)

can anybody please help me out in this question?

 

[jrlaguna]

Hm... I don't have the solution, but can give some idea: The factor decomposition is funny, using 9 instead of 3. Anyway, put it like 9^n2 (7*11)^n1. Now get numbers of the form 4k+1: 1, 5, 9, 13, 17, 21, 25... 81... So, you see: powers of 9 are of the form 4k+1. You have to prove that no number of that form can be a multiple of 7 or 11.