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Define a semi congruent number $\,n\,$ as a congruent number which represents the area of a triangle $\,\bigtriangleup ABC\,$ which has one leg in $\,\mathbb{Q}\setminus \mathbb{Z}\,$ and the other in $\mathbb{Z}.\,$ Are there integer solutions to the equation

$$a^2 + \left(\frac{c}{d}\right)^2= \left(\frac{e}{f}\right)^2$$

with $\,\gcd(c,d)=1=\gcd(e,f)\,$ such that $\,f\neq 1\neq d,\,$ and $\,d\vert a .\quad$ Earlier we discovered there is no non-trivial solutions unless $\,f=d.\quad$ If we restrict $\,d\vert a\,$ are there solutions? If so then the number $\quad n = \dfrac{1}{2} a\cdot \dfrac{c}{d}\quad$ is a semi-congruent number. Otherwise, there is no semi-congruent numbers.

Bill Dubuque
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Roy Burson
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  • So you're looking for a pythagorean triplet where one of the legs has a square term. EG $2^2 +( \frac{3}{2} )^2 = (\frac{5}{2} ) ^2$, $ 6^2 + ( \frac{5}{2} ) ^2 = ( \frac{13}{2} ) ^2$. – Calvin Lin Dec 23 '23 at 22:34
  • this is not semi congruent because the area of such triangle is $n = 1/2 a \dfrac{c}{d}$ which need to be integer. Here we get $n = 1/2 a \dfrac{c}{d} = 3/2$. So we need the number of factors of 2 in the product $ac$ to be larger then number of factors of 2 in $d$ – Roy Burson Dec 23 '23 at 22:40

2 Answers2

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[Pre edit, where OP didn't require that $\frac{1}{2} a \frac{c}{d}$ is an integer.]

With $d=f$, you want $(ad)^2 + (c)^2 = (e)^2$.
With $d \mid a, d \neq 1$, this means that $ad$ must not be square free.
So you're looking for a primitive pythagorean triplet where one of the legs has a square term. With that, we can divide by non-1 factors, to get a solution.

EG Starting with $4^2 + 3^2 = 5^2$, we get $2^2 +( \frac{3}{2} )^2 = (\frac{5}{2} ) ^2$.
Starting with $12^2 + 5^2 = 13^2$, we get $ 6^2 + ( \frac{5}{2} ) ^2 = ( \frac{13}{2} ) ^2$.

[Addressing the edit, where OP requires that $\frac{1}{2} a \frac{c}{d}$ is an integer]

A sufficient condition is $p^2 \mid ad$ where $ p > 2$ is a prime, and we set $ d = p$.

EG Starting with $45^2 + 28^2 = 53^2$, we get $ 15^2 + (\frac{28}{3} )^2 = (\frac{ 53}{3})^2$

Another sufficient condition is that $4^2 \mid ad$ and we set $ d = 2$.

EG Starting with $80^2 + 39^2 = 89^2$, we get $ 40^2 + (\frac{39}{2})^2 = ( \frac{89}{2} ) ^2$.

Calvin Lin
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  • So you shown that $n=60$ is semi-congruent. – Roy Burson Dec 23 '23 at 22:45
  • @Life_is_pie Not sure how you came to that conclusion. Can you elaborate? – Calvin Lin Dec 23 '23 at 22:49
  • The area of right triangle is 1/2 length * width. You provided us with the triple $15^2+ (28/3)^2 = (53/3)^2$. This represents the area with length $15$ and width $28/3$. The area is $60$ and a congruent number. We just shown it is semi-congruent by construction – Roy Burson Dec 23 '23 at 22:52
  • I believe that the area of that triangle is not 60. Can you recheck? – Calvin Lin Dec 23 '23 at 22:54
  • lol it is 70 sorry – Roy Burson Dec 23 '23 at 22:56
  • We can conclude that yes semi congruent numbers exist. – Roy Burson Dec 23 '23 at 22:57
  • Yes, I agree with you now. – Calvin Lin Dec 23 '23 at 22:57
  • Thank you it helped. Although this is one kind of congruent numbers there are others. There are proper ones and improper ones which Euclid found (the primitive triples * constant). To find proper congruent (areas of triangles with legs in strictly $\mathbb{Q}\setminus \mathbb{Z}$) numbers we need to search when both primitive triples are not square free so we can break it up. – Roy Burson Dec 23 '23 at 23:03
  • please see next question for more points :) https://math.stackexchange.com/questions/4832747/does-each-proper-congruent-number-correspond-to-some-pythagorean-triple – Roy Burson Dec 23 '23 at 23:14
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Since $f=d$ and $d$ divides $a$ I would rewrite the equation with $d = xy, a = xz$ :

$$ (xz)^2 + (c/(xy))^2 = (e/(xy))^2 $$

Multiply both sides by $(xy)^2$ :

$$(x^2 yz)^2 + c^2 = e^2 $$

We know for positive integers $m,n,k$ with $m > n$ :

$$(kmn)^2 + (km^2 - kn^2)^2 = (km^2 + kn^2)^2$$

It is not hard to find solutions to

$$x^2 y z = m n k$$

Now if $c = km^2 - kn^2 $ is also even, you have some of your solutions.

Or ofcourse you solve for

$$ x^2 yz = km^2 - kn^2 $$

and $mnk$ being even.

mick
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