a, b, c > 2

Let the perfect square be denoted: S=a^x+b^y+c^z. x y and z must all be even for S to be a perfect square.

Possible values of x are 0, 2, 4 (Total 3)
Possible values of y are 0, 2 (Total 2)
Possible values of z are 0, 2, 4, 6 (total 4)

So total possible factors = 3*2*4 = 24

I actually disagree with this answer. What is there to say that a, b and c are different? If a, b and c all equaled 3. Then n = 3^14 and would be divisible by 8 perfect squares (powers of 3: 0, 2, 4, 6, 8, 10, 12, 14).

Good!!
Only one problem: Why take the perfect square as a^x + b^y + c^z?
Your objection is honoured and the question is given the necessary repair.

from a^4 we get 2 units, from b^3 we get 1 unit and from c^7 we get 3 units.

Can you please explain

The question is asking what are the number of perfect squares which are factors of n. The perfect squares are a^2, a^2, b^2, c^2, c^2, c^2. But, the list also should have combinations of these factors, such as a^2*b^2, a^2*c^2, ....etc
I am not sure I understand the question properly. Is it asking for number of distinct perfect square factors?

Vemuri wrote:The question is asking what are the number of perfect squares which are factors of n. The perfect squares are a^2, a^2, b^2, c^2, c^2, c^2. But, the list also should have combinations of these factors, such as a^2*b^2, a^2*c^2, ....etc
I am not sure I understand the question properly. Is it asking for number of distinct perfect square factors?

Does this mean that if the question read

If n = a^8 × b^6 × c^14, where the three distinct prime numbers a, b, c > 2; what is the number of perfect squares which are factors of n?

5X4X8 = 160

A - 4 units
B - 3 Units
C - 7 Units

(A+1 )* (B+1) * (C+1) = 160?

If yes, thanks so much - you're great. If no, please explain your units theory - its simple and sizzling and sexy for time constraints.