# Thread: Product of perfect square and primes

1. ## Product of perfect square and primes

Let n be an integer and n>0. Prove that n is the product of a perfect square and (possibly zero) distinct prime numbers.

I know that a perfect square is a number that produces a natural number when taking the square root. So 4, 9, 16, etc. are perfect squares.

So I know that a perfect square can always be represented as m^2 for any integer m.

I just don't know how to get to proving that n must be the product of m^2 and a prime p.

Thank you.

2. ## Re: Product of perfect square and primes

hello there
i dont think ur statement is true
take for example 15=3*5
i see no perfect square in 3*5

3. ## Re: Product of perfect square and primes

Originally Posted by letitbemww
Let n be an integer and n>0. Prove that n is the product of a perfect square and (possibly zero) distinct prime numbers.
In the first reply note that $15=1^2\cdot3\cdot 5$ so it is true.

Here is a strong hint.
$36000=2^5\cdot 3^2\cdot 5^3=2^4\cdot 3^2\cdot 5^2(2\cdot 5)$.

Note that $2^4\cdot 3^2\cdot 5^2$ is a square.

4. ## Re: Product of perfect square and primes

I think I figured it out but I'm not sure if it was what you were hinting at, Plato. So because of the Fundamental Theorem of Arithmetic, every integer can be represented as the product of distinct prime numbers. This tells us that the primes are distinct in the product. Also, since 1 is a perfect square factor of every integer, this is true. I also know that the exponent of the prime factors can be represented as 2x+y where x is an integer and y is either 1 or 0 depending on whether the exponent is odd or even. This makes every prime factor be of the form p^(2x+y) = ((p^x)^2)*(p^y) which is a product of a perfect square and a prime.

Is this right?

5. ## Re: Product of perfect square and primes

Originally Posted by letitbemww
I think I figured it out but I'm not sure if it was what you were hinting at, Plato. So because of the Fundamental Theorem of Arithmetic, every integer can be represented as the product of distinct prime numbers. This tells us that the primes are distinct in the product. Also, since 1 is a perfect square factor of every integer, this is true. I also know that the exponent of the prime factors can be represented as 2x+y where x is an integer and y is either 1 or 0 depending on whether the exponent is odd or even. This makes every prime factor be of the form p^(2x+y) = ((p^x)^2)*(p^y) which is a product of a perfect square and a prime. Is this right?
I think you have the right concept. But in case here it is:
$p^{2k}$ is a square.
$p^{2k+1}=p^{2k}(p)$ is a square times a prime.