Every integer can be written as a sum of a square and square free integer

[SNOOTCHIEBOOCHEE]

Homework Statement

show that every positive integer can be written as the product of a square and a square-free integer (an integer that is not divisible by any perfect squares other than one

The Attempt at a Solution

well i can see by example that this works: i.e 60=2²*3*5


Basically we want to show that any integer n>0 can be written as

n=p₁^a₁p₂^a₂ ...p_k^a_k

if a_i is even, then its a perfect square
if a_i is odd, then p_i^a_i-1 is a perfect square and p_i is square free.

I don't know where to go from here... or if what i have said is correct

any help is appreciated

 

[Dick]

The p's are the prime factorization, right? Then you are done. Multiply the perfect squares to get one factor and multiply the square free parts to get the other factor.

 

[SNOOTCHIEBOOCHEE]

Oh really?

so i am completley done with this problem?

but how do i deal with numbers like 6 whose prime factorization is 2*3

 

[Dick]

2 is 2^1. So that falls into your odd case. 2 is square free and 2^(1-1)=1, which is a perfect square, but not a very interesting one.

 

[SNOOTCHIEBOOCHEE]

ok thanks feel free to lock.

 

[HallsofIvy]

oh really?

So i am completley done with this problem?

But how do i deal with numbers like 6 whose prime factorization is 2*3

1= 1² is itself a square. 6= 1*6: a square times a square free number.