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Pythagorean Triangles

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Please help me with the following problem:

a) 3^2 + 4^2 = 5^2
20^2 + 21^2 = 29^2
119^2 + 120^2 = 169^2

To find another such relation, show that if a^2 + (a+1)^2 = c^2, then
(3a+2c+1)^2 + (3a+2c+2)^2 = (4a+3c+2)^2.

(b) If a^2 + (a+1)^2 = c^2, let u=c-a-1 and v=(2a+1-c)/2. Show that v is an integer and that u(u+1)/2 = v^2. This shows that there are infinitely many square triangular numbers.

Please show as much detail as possible.

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This solution is comprised of detailed explanation and step-by-step calculation of the given problems and provides students with a clear perspective of the underlying concepts.

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a) 3^2 + 4^2 = 5^2
20^2 + 21^2 = 29^2
119^2 + 120^2 = 169^2

To find another such relation, show that if a^2 + (a+1)^2 = c^2, then
(3a+2c+1)^2 + (3a+2c+2)^2 = (4a+3c+2)^2.

Solution:
L.H.S.
(3a+2c+1)2 + (3a+2c+2)2
= (3a)2 +(2c)2+(1)2+ 2(3a)(2c)+ 2(2c)(1)+2(1)(3a) + (3a)2 +(2c)2+(2)2 +2(3a)(2c)+ 2(2c)(2)+2(2)(3a)
= 9a2+4c2+1+12ac+4c+6a+9a2+4c2+4+12ac+8c+12a
= 18a2+8c2+24ac+18a+12c+5

R.H.S.
(4a+3c+2)2
= ...

Solution provided by:
Anwesha Ghosh, MSc
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  • MSc, Osmania University
  • MSc, Indian Institute of Technology - Roorkee (I.I.T.-ROORKEE)
  • BSc, Banaras Hindu University
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