Can you find a more golden Pythagorean triangle?

A right triagle with sides x - 1/x and 2 has hypotenuse x + 1/x (where x > 1).

So we need to solve x - 1/x : 2 = (1 + √5)/2

x - 1/x = 1 + √5

x = (1+√5 + √(10 + 2√5)) / 2

According to Wolfram Alpha, the continued fraction expansion of this is

[3; 1, 1, 11, 1, 9, 1, 25, 1, 6, 3, 2, 3, 10, 24, 6, 133, 6, 1, 1, 3, 1, 1, 3, 12, ...]

Letting x be the fraction generated by the first 2, 3, 4, ..., 10 of the numbers in this list,

Then computing (x - 1/x)/2, we get

15 / 8 ≈ 1

45 / 28 ≈ 1.6

3016 / 1863 ≈1.618

7119 / 4400 ≈1.61

700625 / 433008 ≈ 1.6180

424496 / 262353 ≈ 1.618033

569883075 / 352207108 ≈ 1.6180339

614724165 / 379920428 ≈ 1.61803398

29802502503 / 18418959496 ≈ 1.61803398

Where (1 + sqrt(5))/2 = 1.618033988749894848204586834365638117720309179805762862135448...

and the numerators and the denominators of the listed fractions are the sides of a Pythagorean triangle.

Addendum

Just for the heck of it, I let Mathematica compute

x = FromContinuedFraction[{3,1,1,11,1 ,9,1,25,1,6, 3,2,3,10,24 ,6,133,6,1,1 ,3,1,1,3,12}]

x = 4130493785701831 / 1173386725230941

In which case,

(x – 1/x)/2 = 7842071253386625780181726503540 / 4846666576791423668957241552971

Which corresponds to the Pythagorean triangle

(7842071253386625780181726503540,

4846666576791423668957241552971,

9218907660334817612551510249021)

The difference between

7842071253386625780181726503540 / 4846666576791423668957241552971

And (1 + √5)/2 is

1.598299 × 10^(-31)

What is wrong with irrational side lengths? There is indeed a Golden right triangle. The sides are 1 and φ and the hypotenuse is √(φ² + 1)

Edit: if you mean with integral sides you are correct. One can use the continued fraction representation of φ to get as close as one pleases to a golden triangle but with the trade off of increasingly large lengths.