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My question reads: If c^2=ab and (a,b)=1, prove that a and b are perfect squares.

I began my proof by $a=p_1 p_2 \cdots p_n$ and $b=q_1 q_2\cdots q_m$. Then I gave $c$ its own decomposition as well and said $c=s_1 s_2\cdots s_t$.

From there I squares the c and just got the same except now with 2 in the exponent for each s.

I am not too sure how to continue on from there. Would I need to re-index? Also, I am not too sure when to bring in the fact that their gcd is 1.

Michael Hardy
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Sam
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  • @DietrichBurde that does not answer my question because it assumes a and b are prime but I do not in my question – Sam Jan 25 '17 at 00:21
  • @Sam it doesn't assume that $a$ and $b$ are prime, but that $a$ and $b$ are relatively prime, which is a different definition. – Xam Jan 25 '17 at 00:59
  • @Sam No, it assumes $,a,b,$ are relatively prime, the same as your hypothesis that $,(a,b)= 1.\ \ $ – Bill Dubuque Jan 25 '17 at 00:59
  • @Xam okay I see, I do not understand the answer given. Why can i not equal j? – Sam Jan 25 '17 at 03:02
  • @BillDubuque I see now, however I do not understand why p cannot equal q? – Sam Jan 25 '17 at 03:46
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    Then $,p=q,$ would be a common factor of $,a,b,,$ contra hypothesis. – Bill Dubuque Jan 25 '17 at 04:02
  • @BillDubuque okay because of the gdc=1. So after I have all my prime factorizations written out I can just use this gcd fact to state that a ans b must be square? Do I not have to show divisibility at some point? – Sam Jan 25 '17 at 04:06

2 Answers2

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Pick a prime $p$ that divides $a$, we must show the maximum power of $p$ that divides $a$ is even. But this number is equal to the maximum power of $p$ that divides $c$ (because $c=ab$ and $p$ does not divide $b$), which is even because $c$ is a square.

Asinomás
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Hint $ $ Either compare the parity of the exponents in prime factorizations, or induct as below.

Induct on $\,k =\,$ number of prime factors of $\,c.\,$ If $\,k=0\,$ then $\,c=1\,$ so $\,a=1=b\,$ are squares. Else $\,k\ge 1,\,$ so some prime $\,p\mid c,\,$ so $\,p^2\mid c^2\! = ab\,$ thus $\,p^2\mid a\,$ or $\,p^2\mid b\,$ by $\,a,b\,$ coprime. Wlog $\,p^2\!\mid b,\,$ so canceling $\,p^2$ we get $\,a(b/p^2) = (c/p)^2.\,$ $\,c/p\,$ has less prime factors than $\,c\,$ hence by induction we deduce that $\, a = j^2,\,\ b/p^2 = k^2,\,$ so $\,b = (pk)^2$.

Remark $\ $ The proof generalizes to higher powers and to any UFD. Alternatively a simple proof using gcd laws give an explicit representation, viz. $\ a = \gcd(a,c)^2,\,\ b = \gcd(b,c)^2.$

Community
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Bill Dubuque
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