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I want to show every congruent number $\alpha $ be written as $$\alpha = \dfrac{nm(n-m)(m+n)}{\delta}$$ for some $m,n$ where $\gcd(m,n)=1$, $m\not \equiv n\mod 2 $, $m>n$ and $\delta$ a divisor of $mn$ or $(m-n)(m+n)$. I would also like to show that whenever $\delta$ is a divisor of the square part of $nm(n-m)(m+n)$ then $$\alpha = \dfrac{nm(m-n)(m+n)}{\delta}$$ is a congruent number.

Roy Burson
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1 Answers1

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Using Euclid's formula $\quad A=m^2-n^2\quad B=2mn\quad C=m^2+n^2\quad$ area is by definition $\quad\alpha=\dfrac{2mn(m^2-n^2)}{2} =mn(m-n)(m+n).\quad$ Most congruent numbers can be written with $\,m,n\in\mathbb{N}\quad$ e.g $\quad\alpha=6\implies(m,n)=(2,1)\,$ but some $\quad \bigg(\dfrac{20}{3}\bigg)^2 + \bigg(\dfrac{3}{2}\bigg)^2 =\bigg(\dfrac{41}{6}\bigg)^2 \implies\alpha=5\,$ are elusive: $\quad \alpha=5\implies m\ge2 \sqrt{\dfrac{5}{3}},\, n = \pm \sqrt{m^2 - \dfrac{20}{3}}.\quad$ This means that, while $\,\alpha=5\,$ is congruent number, being an integer area of a right triangle, it is not the area of any Pythagorean triple which requires integers for all sides.

Therefore, Euclid's formula and the area-derived formula you supplied cannot be used to describe all congruent numbers.

mick
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poetasis
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  • here $\delta = 1$ – Roy Burson Dec 24 '23 at 03:39
  • and 1 is perfect square 1*1 = 1 – Roy Burson Dec 24 '23 at 03:40
  • @Life_is_pie The sides of the triangle are not integers for $\alpha=5$ so the formula cannot generate it using integers for $m,n$, Integers are required for divisibility. – poetasis Dec 24 '23 at 03:44
  • where is flaw then because $f=bd$ is true. – Roy Burson Dec 24 '23 at 03:44
  • so it is true for congruent numbers with corresponding triangle whose legs are not integers. – Roy Burson Dec 24 '23 at 03:46
  • Choose $m=5$ and $n=4$ Then $a = (m-n)(m+n)=1\cdot 9=9$, $b = 2\cdot m\cdot n= 2\cdot 3\cdot 2 = 12$. Both $a$ and $b$ are not square free so we can break them up by factoring the largest non square free part of both $a$ and $b$ as $a=3\cdot 3=9$ and $b=2\cdot 20 $. Write $c^2 = (9)^2+(40)^2 = (3\cdot3)^2 +(2\cdot 20)^2$ then it follows that the integer $n = \frac{1}{2}(\frac{3}{2})(\frac{20}{3}) =5$ is a congruent number generated by the triple $(\frac{3}{2},\frac{20}{3}, \frac{41}{6})$. This is precisely the triple that Fibonacci discovered in 1220 in response to his challenger. Here $f=bd$ – Roy Burson Dec 24 '23 at 03:53
  • 12 was suppose to be 20 and then $mn(m-n)(m+n) = 9 20$ which is divisible by 5 so we have $mn(m-n)(m+n)/ 36 = 5$ with $m=5$ and $n=4$ – Roy Burson Dec 24 '23 at 04:02