How to determine pythagorean triples that have a slope closest to 1

Question

I'm not a mathematician, and I'm not sure how to phrase this question properly, so please bear with me as I stumble through the question.

Considering Pythagorean's Theorem a²+b²=c²

I'm looking for solutions that fit the following requirements:

1. two of the prime leg, even leg or hypotenuse must be prime numbers
2. the third number must be an integer
3. the slope is close to 1

The third requirement is obviously not clear. Here are some examples:

prime leg   even leg   hypotenuse   slope
---------   --------    ---------   -----
      3          4            5     0.750
      5         12           13     0.417
     11         60           61     0.183
     19        180          181     0.106
     29        420          421     0.069
     59       1740         1741     0.034
     61       1860         1861     0.033

But as you see in the table, the larger numbers become less and less steep.

How would you go about calculating this?

Answer 1

You are going to have a hard time with this. Clearly, your Pythagorean triples are primitive. Which means that there are natural numbers $u,v$ such that your so-called prime leg is equal to $u^2-v^2$, the even leg is equal to $2uv$ and the hypotenuse is equal to $u^2+v^2$.

The fact that your prime leg is equal to $u^2-v^2=(u-v)(u+v)$, and at the same time a prime number means that we must have $u-v=1$. Rewriting the above expressions for the three sides using this gives a prime leg of length $2v+1$ and even leg of length $2(v^2+v)$. The ratio between these two sides is just going to get further and further away from $1$ as you pick bigger and bigger primes for your prime side.

As a side note, the length of the hypotenuse will always be $(v+1)^2+v^2=2(v^2+v)+1$, which is $1$ more than the length of the even leg. You probably already spotted this pattern from your table.

Answer 2

I can't help much with the prime numbers but I can help with getting close to a slope of $1$ in Pythagorean triples. We want to find triples where $|A-B|=1$ and, to do that, we solve $A^2+(A\pm1)^2=C^2$ for $n$. (I'll spare you the details.) These numbers get rarer and rarer so most can only be found programmatically.

$$\text{We let }(n=\sqrt{2m^2\pm1}-m)\text{ and select }m,n\text{ whenever n is an integer and }n<m.\text{ We use }\pm\text{in our test because sometimes }A>B\text{ and sometimes }A<B.$$ Then we can generate the following triples as $f(m,n)$ using Euclid's formula. $$f(2,1)=(3,4,5)\quad 0.75$$ $$f(5,2)=(21,20,29)\quad 1.05$$ $$f(12,5)=(119,120,169)\quad 0.991666666666667$$ $$f(29,12)=(697,696,985)\quad 1.0014367816092$$ $$f(70,29)=(4059,4060,5741)\quad 0.999753694581281$$ $$f(169,70)=(23661,23660,33461)\quad 1.00004226542688$$ $$f(408,169)=(137903,137904,195025)\quad 0.999992748578721$$ $$f(985,408)=(803761,803760,1136689)\quad 1.00000124415248$$ $$f(2378,985)=(4684659,4684660,6625109)\quad 0.999999786537337$$ $$f(5741,2378)=(27304197,27304196,38613965)\quad 1.00000003662441$$ $$f(13860,5741)=(159140519,159140520,225058681)\quad 0.999999993716245$$ $$f(33461,13860)=(927538921,927538920,1311738121)\quad 1.00000000107812$$ $$f(80782,33461)=(5406093003,5406093004,7645370045)\quad 0.999999999815024$$ $$f(195025,80782)=(31509019101,31509019100,44560482149)\quad 1.00000000003174$$ $$f(470832,195025)=(183648021599,183648021600,259717522849)\quad 0.999999999994555$$ $$f(1136689,470832)=(1070379110497,1070379110496,1513744654945)\quad 1.00000000000093$$ $$f(2744210,1136689)=(6238626641379,6238626641380,8822750406821)\quad 0.99999999999984$$ $$f(6625109,2744210)=(36361380737781,36361380737780,51422757785981)\quad 1.00000000000003$$ $$f(15994428,6625109)=(211929657785303,211929657785304,299713796309065)\quad 0.999999999999995$$