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I need to prove that for every integer $a>0$ there is a unique representation $a=r*s^2$ where $r$ is not dividable by any square: there is no $d>1$ such that $d^2|r$

What I tried is to show a as a unique multiplication of primes and then show the case that a is odd or not, but didn't get anywhere.. any help will be appriciated

Martin Sleziak
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user2993422
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    Write $a=p_1^{e_1}\cdots p_r^{e_r}$ with primes $p_i$, and consider the exponents (even and odd). – Dietrich Burde Jun 09 '15 at 13:53
  • Have you tried finding such a representation on actual integers (like say $3$, $4$, $5$, $\ldots$ $12$, $\ldots$ $17$, $18$, $19$, $20$), to get a sense of what is going on ? – Joel Cohen Jun 09 '15 at 13:55

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You don't have to separate the cases when $r$ is odd. Let me show you how you can do that for a particular $a$, you will probably get the idea:

$$360 = 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5 = (2^2 \cdot 3^2) \cdot (2\cdot 5)$$

So we can take $r=10$ and $s=6$.

5xum
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