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when generating primitve Pythagorean triples and building sorted groups that contain a cathetus of specific length, I saw in one group an exceptionally high jump in the lengths of the second cathetus in that group:

Exceptional jump in cathetus length

while the observed usual growth factor is below 2, that factor is about 127 from the highlighted triple to the next one with common cathetus of length 1021020.

Questions:

  • has this exceptional behavior been noticed before
  • are there any upper bounds on the quotient of "adjacent" catheti-lenghts in the sorted groups with common cathetus-length
  • will the jumps be observed eventually for every cathetus-length in the enumeration of all primitive Pythagorean triples
  • do the triples for which these jumps are observed have special properties; in the example the difference between the common cathetus-length and the hypotenuse is $1$

$$\,$$
here is list of primitive Pythagorean triples where the the length of the maximal cathetus and of the hypotenuse differ by 1, grouped according to common cathetus length. To be able to compare ratios within a group, only groups with 3 Pythogeran triples are listed:

[(60, 11, 61), (60, 91, 109), (60, 221, 229), (60, 899, 901)]
[(84, 13, 85), (84, 187, 205), (84, 437, 445), (84, 1763, 1765)]
[(180, 19, 181), (180, 299, 349), (180, 2021, 2029), (180, 8099, 8101)]
[(220, 21, 221), (220, 459, 509), (220, 3021, 3029), (220, 12099, 12101)]
[(264, 23, 265), (264, 1073, 1105), (264, 1927, 1945), (264, 17423, 17425)]
[(312, 25, 313), (312, 1505, 1537), (312, 2695, 2713), (312, 24335, 24337)]
[(364, 27, 365), (364, 627, 725), (364, 8277, 8285)]
[(420, 29, 421), (420, 341, 541), (420, 851, 949), (420, 1189, 1261), (420, 1739, 1789), (420, 4891, 4909), (420, 11021, 11029)]
[(480, 31, 481), (480, 2279, 2329), (480, 6391, 6409)]
[(612, 35, 613), (612, 1075, 1237), (612, 23405, 23413)]
[(760, 39, 761), (760, 5751, 5801), (760, 9009, 9041)]
[(840, 41, 841), (840, 559, 1009), (840, 1081, 1369), (840, 3551, 3649), (840, 7031, 7081), (840, 11009, 11041), (840, 19591, 19609)]
[(924, 43, 925), (924, 893, 1285), (924, 1643, 1885), (924, 4307, 4405), (924, 5893, 5965), (924, 23707, 23725)]
[(1740, 59, 1741), (1740, 3139, 3589), (1740, 7469, 7669), (1740, 20989, 21061)]
[(1860, 61, 1861), (1860, 3619, 4069), (1860, 8549, 8749), (1860, 23989, 24061)]
[(2244, 67, 2245), (2244, 2117, 3085), (2244, 4067, 4645), (2244, 10283, 10525)]
[(2380, 69, 2381), (2380, 4611, 5189), (2380, 7029, 7421), (2380, 14061, 14261)]
[(2520, 71, 2521), (2520, 1241, 2809), (2520, 3569, 4369), (2520, 19519, 19681)]
[(2964, 77, 2965), (2964, 2573, 3925), (2964, 5723, 6445), (2964, 12827, 13165)]
[(3120, 79, 3121), (3120, 3649, 4801), (3120, 10591, 11041), (3120, 14231, 14569)]
[(3444, 83, 3445), (3444, 6283, 7165), (3444, 14933, 15325)]
[(3612, 85, 3613), (3612, 6955, 7837), (3612, 16445, 16837)]
[(3960, 89, 3961), (3960, 1729, 4321), (3960, 9401, 10201)]
[(4140, 91, 4141), (4140, 7571, 8629), (4140, 12901, 13549)]
[(5100, 101, 5101), (5100, 4469, 6781), (5100, 9779, 11029), (5100, 22211, 22789)]
[(5304, 103, 5305), (5304, 3103, 6145), (5304, 24047, 24625)]
[(5940, 109, 5941), (5940, 11371, 12829), (5940, 17741, 18709)]
[(6160, 111, 6161), (6160, 4329, 7529), (6160, 6519, 8969)]
[(6384, 113, 6385), (6384, 17113, 18265), (6384, 22663, 23545)]
[(6612, 115, 6613), (6612, 6125, 9013), (6612, 12155, 13837)]
[(8580, 131, 8581), (8580, 3059, 9109), (8580, 10579, 13621), (8580, 15811, 17989), (8580, 19549, 21349)]
[(9660, 139, 9661), (9660, 8909, 13141), (9660, 11461, 14989), (9660, 17819, 20269)]
[(11100, 149, 11101), (11100, 9821, 14821), (11100, 21131, 23869)]
[(12012, 155, 12013), (12012, 3925, 12637), (12012, 18685, 22213), (12012, 22195, 25237)]
[(14280, 169, 14281), (14280, 6401, 15649), (14280, 10561, 17761), (14280, 16999, 22201)]
[(16380, 181, 16381), (16380, 8789, 18589), (16380, 11651, 20101), (16380, 12931, 20869)]

Improve this question
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5
  • $\begingroup$ Does this happens every time you considers triplets of the form $m$, $(m^2-1)/2$, $(m^2+1)/2$ ? $\endgroup$ Commented 2 days ago
  • $\begingroup$ @G.Melfi yes, that seems tobe the case $\endgroup$ Commented 2 days ago
  • 2
    $\begingroup$ The Euclidean parameterisation for primitive triples is $(v-u)(u+v), 2uv, u^2+v^2$ where $\gcd(u,v)=1$. If we instead take $v = u+d$ with $\gcd(u,d) = 1$ we get $(2ud+d^2, 2ud+2u^2, 2ud+2u^2+d^2)$. This has potential with small $d$, because most numbers with one factor just smaller than the square root don't have a second such factor. You've considered $d=1$ (and according to my calculations $b=112$ gives ratio $209$, $b=1984$ gives ratio $\sim 15620$) but other small $d$ also give good ratios (e.g. $b=7807$ with $d=3$ and ratio $\sim 40643$). $\endgroup$ Commented 2 days ago
  • $\begingroup$ The Pythagorean triples that share a leg can be obtained separating the factors of the square of the leg in two groups. All factors corresponding to the same prime must go to the same side, except for the case of $2$ (one $2$ must go to one side and all the others to the other side). Then, the more similar is the product, the smaller is the resulting leg. In this case, $1021020=2^2\times3\times5\times7\times11\times13\times17$. The smallest half difference is $(5\times11\times13)^2-(2\times3\times7\times17)^2=1429$, and the next one is $(2\times5\times7\times11)-(3\times13\times17)^2=153331$. $\endgroup$ Commented 2 days ago
  • 1
    $\begingroup$ So, there are multiple ways of getting high quotients. For a leg that is the product of two similar odd primes the quotient is $\frac{p^2q^2-1}{p^2-q^2}$, that grows fast and is not bound. For instance, for $10403$ the quotient is $265251$. For a leg that is a power of $2$ multiplied by a prime number, like $1984$ in @PeterTaylor's response, the quotient is $\frac{2^{2k-2}p^2-1}{2^{2k-2}-p^2}$. It is very high when the square of the prime is similar to a power of $2$. For instance, the quotient is $\sim 1036304$ for $32512$ $(k=8, p=127)$. $\endgroup$ Commented 2 days ago

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