# Integer as a product of a square and a square free integer

My question actually relates to an example given on p. 28 of Julian Havil's "Gamma".

Discussing a proof of the infinity of primes due to Erdos, Havil writes:

[Erdos] uses a counting technique ... used by number theorists: that any integer can always be written as the product of a square and a square-free integer...

And then gives, as an example:

2851875 = 3^3 x 5^4 x 11 x 13^2 = 3 x 11 x (3 x 5^2 x 13)^2


But how can

(3 x 5^2 X 13)^2 = 3^2 x 5^4 x 13^2


?

Oops I should pay more attention - deleting this shortly, thanks for help in pointing out stupidity

Further Actually cannot delete as there is an answer so this monument to my inability to tell the difference between addition and multiplication will have to stay here!

• oh I have been a bit of an idiot! Oh dear – adrianmcmenamin Aug 4 '14 at 11:30

If an integer is already a perfect square, then $n^2=n^2\cdot 1$
• But $1$ is not square-free. ;-) – Lucian Aug 4 '14 at 15:30