Foo numbers (using recursion)

[galaxy]

Hi guys

I have this small assignment to do about foo numbers
that I should implement using a recursive.

First of all, foo numbers don't exist
Problem statement:
We define the concept of a foo number as follows. 0 is a foo number. If i is a foo number then i+5, and i+7 are also foo numbers. Write a recursive method which will test if a given integer is a foo number.

My code:

#include <iostream>
using namespace std;

bool isFoo(int n);

void main(){
	int n;

	cout<<"Enter a number"<<endl;
	cin>>n;

	while(cin){
		if(isFoo(n))
			cout<<n<<" is a Foo number"<<endl;
		else
			cout<<n<<" is not a Foo number"<<endl;
		cout<<"Enter a number"<<endl;
		cin>>n;
	}

}

bool isFoo(int n){
	if(n == 0)
		return true;
	else if(n < 0)
		return false;
	else
		return false || isFoo(n - 5) || isFoo(n - 7);
}

The idea of the question is that you return true if the number is a multiple of
5 or 7, but I should implement it using recursive function.

The code works for both multiples of 5 and 7 but there's a logical
error that I can't figure out when you enter 12 it prints 12 is a Foo number
because at a certain case when the isFoo(n-5) returns, the value of n will be
7 and isFoo(n-7) will be true;

I can't find a way to do each term individually
I mean I want to find a way
that I keep subtracting the number by 5 untill we get either zero or negative
if it's zero then it's a foo number otherwise

keep subtracting the number by 7 untill
we get either zero or negative
if it's zero then it's a foo number otherwise it's not

do anybody have an idea for solving this problem
I'm new at recursion and I will be grateful for your help

[Duthomhas]

You don't have to use only one function to define foo.

:-)

[galaxy]

actually it was written in the problem statement but I forgot to copy it
he already defined the header of the function to be:
bool isFoo(int n)

[Bazzy]

---Don't care about this post---

[Gumbercules]

12 is a foo number, it's not just multiples 0 + 5 + 7 = 12