Primitive Pythagorean triples with the odd leg a Pentagonal number

 
 
There are infinitely many primitive Pythagorean triples   (a, b, c),   like   (5, 12, 13),
with hypotenuse c such that the odd leg is a pentagonalnumber and the even leg is consecutive with the hypotenuse.

The odd leg a being the pentagonal number   a_n = n \,(3 \,n - 1)/2
The even leg   b = 2 \,k
The hypotenuse   c = 2 \,k + 1

The odd pentagonal numbers   a_n   can be partitioned into two sets by

n = 2 + 4 \, p   with   p = 0, 1, 2,...
n = 5 + 4 \, m   with   m = 0, 1, 2,...
 

If   n = 2 + 4 \, p,   then

a_n = n \,(3 \, n - 1)/2 = (2 + 4 \, p)(3 \, (2 + 4 \, p) - 1)/2 = 24 \, p^2 + 22 \, p + 5

a^2 + b^2 = c^2

(24 \, p^2 + 22 \, p + 5)^2 + 4 \, k^2 = (2 \, k + 1)^2
(24 \, p^2 + 22 \, p + 5)^2 = (2 \, k + 1)^2 - 4 \, k^2
576 \,p^4 + 1056 \,p^3 + 724 \,p^2 + 220 \,p + 25 = 4 \, k + 1

k = 144 \, p^4 + 264 \, p^3 + 181 \, p^2 + 55 \, p + 6

Since the equation has a positive integer solution for   p = 0, 1, 2,...
there are infinitely many Pythagorean triples of the desired form.

a = 24 \, p^2 + 22 \, p + 5
b = 2 \,(144 \, p^4 + 264 \, p^3 + 181 \, p^2 + 55 \, p + 6)
c = 2 \,(144 \, p^4 + 264 \, p^3 + 181 \, p^2 + 55 \, p + 6) + 1

 

Here are the first few Primitive triples:

(5,12,13) ………………                5 = P_2
(51,1300,1301) ………….          51 = P_6
(145,10512,10513) ……….      145 = P_{10}
(287,41184,41185) ……….      287 = P_{14}
(477,113764,113765) ……..     477 = P_{18}
(715,255612,255613) ……..     715 = P_{22}
(1001,501000,501001) …….   1001 = P_{26}
(1335,891112,891113) …….   1335 = P_{30}
(1717,1474044,1474045) ….. 1717 = P_{34}
(2147,2304804,2304805) ….. 2147 = P_{38}
(2625,3445312,3445313) ….. 2625 = P_{42}
……………………
……………………

 

If   n = 5 + 4 \, m,   then

a_n = n \,(3 \, n - 1)/2 = (5 + 4 \, m)(3 \, (5 + 4 \, m) - 1)/2 = 24 \, m^2 + 58 \, m + 35

(24 \, m^2 + 58 \, m + 35)^2 = 4 \, k + 1

k = 144 \, m^4 + 696 \, m^3 + 1261 \, m^2 + 1015 \, m + 306

there are infinitely many Pythagorean triples of the desired form.

a = 24 \, m^2 + 58 \, m + 35
b = 2 \,(144 \, m^4 + 696 \, m^3 + 1261 \, m^2 + 1015 \, m + 306)
c = 2 \,(144 \, m^4 + 696 \, m^3 + 1261 \, m^2 + 1015 \, m + 306) + 1

 

Here are the first few Primitive triples:

   (35,612,613) ……………          35 = P_5
   (117,6844,6845) …………       117 = P_9
  (247,30504,30505) ……….      247 = P_{13}
  (425,90312,90313) ……….      425 = P_{17}
  (651,211900,211901) ……..     651 = P_{21}
  (925,427812,427813) ……..     925 = P_{25}
(1247,777504,777505) …….    1247 = P_{29}
(1617,1307344,1307345) …..   1617 = P_{33}
(2035,2070612,2070613) …..   2035 = P_{37}
(2501,3127500,3127501) …..   2501 = P_{41}
(3015,4545112,4545113) …..   3015 = P_{45}
………………
………………

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged , . Bookmark the permalink.

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