(1+3+5+.....+p)+(1+3+5........+q)=(1+3+5+.....+r)

where p q r are real p greater than 6

what is the smallest possible value of p+q+r?

1+3+5+......p=p/2[p+(p-1)2]

1+3+5+......q=q/2[q+(q-1)2]

1+3+5+......r=r/2[r+(r-1)2]

p/2[p+(p-1)2] + q/2[q+(q-1)2] = r/2[r+(r-1)2]

On simplification,

p² + q² = r²

Given:

p>6 ; On further simplification,

we get p=8 , q=6 , r=10 are the smallest values

=> p + q + r = 24

actually the answer is cumin 21

(book: tata mc graw hill iit jee - 2010)