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Define a proper congruent number as an integer which represents the area of some right triangle $\triangle ABC$ whose legs are both strictly in $\mathbb{Q}/\mathbb{Z}$. For every proper congruent number there exist $a$, $b$, $c$, $d$, $e$, and $f$ such that

\begin{equation} \left(\dfrac{a}{b}\right)^2 + \left(\dfrac{c}{d}\right)^2 = \left(\dfrac{e}{f}\right)^2, \end{equation}

with $d\mid a $ and $b\mid c$. Is it true that $f = bd$ and $e^2 = (ad)^2+(bc)^2$? If so we see that each proper congruent number corresponds to some Pythagorean triple that has non square free components.

Xander Henderson
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Roy Burson
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  • Do you also want $ac/(2bd)$ to be an integer? – Calvin Lin Dec 23 '23 at 23:22
  • This is your 3rd question about pythagorean triples today, and all very similar. You got helpful answers. I suggest you try this one yourself. Show some effort in your question or give an answer. – mick Dec 23 '23 at 23:23
  • @mick there you go – Roy Burson Dec 23 '23 at 23:24
  • Why edit the MathJax into an identical looking expression? – Тyma Gaidash Dec 23 '23 at 23:28
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    @ТymaGaidash If you are asking about my edit, it doesn't look identically the same. $\bigtriangleup ABC$ is not the same as $\triangle ABC$, and $a\vert b$ is not the same as $a \mid b$. In both cases, the spacing in the former is not quite correct, and the spacing in the latter is better. – Xander Henderson Dec 23 '23 at 23:58
  • Note $,b\mid c,\ (c,d)=1\Rightarrow (b,d)=1,$ so $,(b^2,d^2)=1,$ so the denominator of the sum $ = {\rm lcm}(b^2,d^2)=b^2d^2,$ by here, so the denoms are equal $,f^2 = b^2d^2,$ by unique fractionization – Bill Dubuque Dec 24 '23 at 00:08
  • @BillDubuque I dont think we need unique factorization just show both sides are in lowest terms – Roy Burson Dec 24 '23 at 00:19
  • @BillDubuque if you have two fractions in $\mathbb{Q}$ that are in lowest terms then we can show that there sum of squares are also in lowest terms. So reduces rational numbers are closed under this operation. – Roy Burson Dec 24 '23 at 00:20
  • Euclid's Lemma, gcd laws, etc are all equivalent to uniqueness of prime factorizations. Uniqueness of reduced fractions easily follows from such (and is equivalent to unique factorization). Using uniqueness of reduced fractions is usually the easiest way to prove results like this (e.g. said one-line proof here). – Bill Dubuque Dec 24 '23 at 00:26
  • @BillDubuque thanks so it does mean that each proper congruent number corresponds to some primitive triple $(a,b,c)$ right? We should actually go backwards and say if $e^2 = (ad)^2+(bc)^2$ for some integers then $n= 1/2\dfrac{ac}{bd}$ is a proper congruent. – Roy Burson Dec 24 '23 at 00:31
  • That the least denom of the sum of two reduced fractions with coprime denoms is the product of their denoms (cf. my prior comment) is proved here in many places, e.g. see here in the 2nd linked dupe. – Bill Dubuque Dec 24 '23 at 00:44
  • @BillDubuque thanks that will shorten what I need. If $(ad ,bc,e) $ is a primitive triple with $a\vert d$ and $b\vert c$ then $n=1/2\dfrac{ad}{bc}$ is a congruent number when $ad$ containts more factors of 2 then $bc$. Similarly, if $n$ is a proper congruent then $e^2 = (ad)^2 + (bc)^2$ with $a\vert d$ and $b\vert c$ . So there seem to be mapping from each primitve Pythagorean triple to each congruent number. – Roy Burson Dec 24 '23 at 00:50

1 Answers1

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Suppose $n$ is a proper congruent number with $1=\gcd(a,b)=\gcd(c,b)=\gcd(e,f)$. Since $n$ is congruent it represents the area of the triangle $\triangle ABC$ which has area $n=\frac{1}{2}(\frac{a}{b})(\frac{c}{d}) = \frac{1}{2}\frac{ac}{bd}$. Since $n$ is an integer it must follow that $bd\vert ac$. Since $n$ is proper we know that $b$ must divide $c$ and $d$ must divide $a$. Let $i$ denote the largest power of $b$ so that $b^{i}\vert c$ but $b^{i+1} \not \vert c$. Similarly, let $j$ denote the largest power of $d$ so that $d^{j}\vert a$ but $d^{j+1} \not \vert a$. Write $a =d^{j}q_2$ and $c =b^iq_1$ for some quotients $q_1$ and $q_2$, then by the hypothesis it follows that

\begin{equation} \label{Equation 1} \begin{split} (\frac{a}{b})^2+(\frac{c}{d})^2 & =(\frac{e}{f})^2\\ & \Downarrow\\ (\frac{d^{j}q_2}{b})^2+(\frac{b^iq_1}{d})^2 & =(\frac{e}{f})^2\\ & \Downarrow\\ \frac{d^{2j}q_2^2}{b^2}+\frac{b^{2i}q_1^2}{d^2} & =\frac{e^2}{f^2}\\ & \Downarrow\\ \frac{d^{2j+1}q_2^2+ b^{2i+1}q_1^2}{b^2d^2} & =\frac{e^2}{f^2}\\ \end{split}\tag{1} \end{equation}

Now suppose that the numerator and the denominator of the last equality in Eq.(1) are not in lowest terms. Then it must follow that $$b^2d^2\vert \left(d^{2j+1}q_2^2+ b^{2i+1}q_1^2\right)$$ but $bd\vert b^2d^2$ so then $$bd\vert b^2d^2\vert \left(d^{2j+1}q_2^2+ b^{2i+1}q_1^2\right).$$ Thus the product $bd$ divides the numerator. This implies $b$ and $d$ divide the numerator individually, but this cannot happen because $b$ divides the numerator if and only if $b\vert q_2^2$ (as $\gcd(b,d) = 1$ as $d$ is a divisor of $a$), but $a=d^jq_2$ and $1=\gcd(a,b)=\gcd(b,d^jq_2)$, which means $b$ cannot divides $q_2$. The same type of argument holds for the integer $d$. Thus the numerator and the denominator are in lowest terms, which implies that is $f^2 = b^2d^2 = (bd)^2$. Hence, $f=bd$ as desired.

Roy Burson
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