2010/048) given rational numbers p, q , r satisfying

pq+qr+rp=1

prove that (psquare +1)(qsquare +1)(rsquare +1) is the square of a rational no.

proof:
pq+qr+rp = 1

so p = (1-qr)/(q+r)

if we chose q = tan A and r = tan B

we get

1/p = (q+r)/(1-qr) = (tan A + tan B)/(1- tan A tan B) = tan (A+B)

or p = cot (A+B)

we can chose q and r to be <1 -ve="" also="" br="" case="" else="" even="" in="" positive="" want="" we="">(p^2+1) (q^2+1)(r^2+ 1) = sec^2 A sec ^2 B cosec^2 (A+B)


= (sec^2 A sec^B)/ sin^2 (A+B)

this is square of reciprocal of sin (A+B) cos A cos B
sin (A+B) cos A cos B
=( sin A cos B + cos A sin B)cos A cos B
= sin A cos A cos ^2B + cos^2 A sin B cos B
= tan A cos ^2 A cos ^2 B + tan B cos ^2 A cos ^2 B
= (tan A + tan B)/(sec^2 A sec ^2B)
= (tan A + tan B)/(1+ tan ^2 A)(1+ tan ^2B)

so (p^2+1) (q^2+1)(r^2+ 1) = ((1+ tan ^2 A )(1+tan ^2B)/(tan A + tan B))^2

if tan A and tan B that is q and r are rational then

((1+ tan ^2 A )(1+tan ^2B)/(tan A + tan B)) is rational and so (psquare +1)(qsquare +1)(rsquare +1) is the square of a rational no.