The nth triangular number T n is given by the formula T n = 1 + 2 +3 +...+n = (n

Question

The nth triangular number Tn is given by the formula Tn = 1 + 2 +3 +...+n = (n(n+1))/2. The first few triangular numbers are 1, 3, 6, and 10. In the list of the first few Pythagorean triples (a, b, c), we find (3, 4, 5), (5, 12, 13), (7, 24, 25), and (9, 40, 41). Notice that in each case, the value of b is four times a triangular number. If you believe that this is true, then prove it. Otherwise find some triangular number for which it is not true.

a) Find a primitive Pythagorean triple (a, b, c) with b= 4T5 . Do the same for b= 4T6 and for b= 4T7
b) Do you think that for every triangular number Tn , there is a primitive Pythagorean triple (a, b, c) with b= 4Tn . If you believe that this is true, then prove it. Otherwise find some triangular number for which it is not true.

Explanation / Answer

a) b=4T5 T5=15 b=60 the triplet is (11,60,61) b=4T6=84 the triplet is (13,84,85) b=4T7=112 the triplet is (15,112,113) b) for every Tn b=4Tn there is a triplet (2n+1,2n(n+1),2n(n+1)+1) by considering Tn=n(n+1)/2 we get b=2n(n+1) now considering c=(2n(n=1)+1 therefore (a,b,c) is to be a pythogoreans triplet... it should obey a^2+b^2=C^2 using this we get the value of a as follows we get a=2n+1

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