Thread: n

**mancillaj3** · February 14th 2009, 12:36 AM

Prove that:

if a and b are positive integers with (a, b) = 1, and if ab is a perfect square,

then a and b are perfect squares.

**red_dog** · February 14th 2009, 06:35 AM

Let $ab=k^2$

Let $k=p_1p_2\ldots p_n$ . Then $ab=p_1^2p_2^2\ldots p_n^2$

Then $p_1^2|ab$ . But $(a,b)=1\Rightarrow p_1^2|a$ or $p_1^2|b$ . Suppose that $p_1^2|a$

In the same way, suppose $p_2^2|a,\ldots ,p_i^2|a$ .

Then $p_{i+1}^2,\ldots ,p_n^2|b$

So, $a=p_1^2p_2^2\ldots p_i^2a_1, \ b=p_{i+1}^2p_{i+2}^2\ldots p_n^2b_1$

Then $ab=p_1^2p_2^2\ldots p_n^2a_1b_1\Rightarrow a_1=b_1=1\Rightarrow a=(p_1p_2\ldots p_i)^2, \ b=(p_{i+1}p_{i+2}\ldots p_n)^2$

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