Let (a/x)^2+(b/y)^2=c, where the rational fractions cannot be further resolved. It's easily shown that a^2+b^2=cx^2, for some integer, x (see, e.g. xdogs' first post).
But if so, then x = (a^2+b^2)^(1/2)/c^(1/2), but since we are dealing with square roots, once the fraction is again fully resolved, this can be true only if c=a^2+b^2, i.e. x=1, or a^2+b^2=x^2, i.e. c=1.
In either case, it seems that c is the sum of rational squares iff it is the sum of integer squares, a non-trivial result.
Since 14 is not a sum of integer squares, the solution of the problem follows at once.
Edited on May 8, 2015, 10:25 am
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Posted by broll
on 2015-05-08 09:15:06 |
Generalisation.
