David sent in this solution, using the
hints we gave you.
Öa+Öb rational
rational
rational
rational
rational
i.e., Öa(Öa+Öb) rational
Þ Öa rational (and so a is a square)
Þ Öb is also rational and hence b is a square.
For the second part:
Öa+Öb+Öc rational
Þ (Öa+Öb+Öc)2 rational
| Þ |
| __
Öa b
|
+ |
| __
Öb c
|
+ |
| __
Öc a
|
|
rational
| ( |
| __
Öa b
|
+ |
| __
Öb c
|
+ |
| __
Öc a
|
)2=a b+b c+c a+2 |
| ____
Öa b c
|
(Öa+Öb+ Öc)
|
so
is also rational
| Þ Öa( |
| __
Öa b
|
+ |
| __
Öb c
|
+ |
| __
Öc a
|
)- |
| ____
Öa b c
|
=a(Öb +Öc)
|
is rational
so Öb+Öc is rational and Öa is rational, and use the above.