**Contribution
for Puzzlet #071**
**From: Ken Duisenberg [****ken.duisenberg@hp.com]**

From
the problem statement, we know that

c = -(92^2-b^2)/4a = ([b/2]^2 - 46^2)/a
For
c to be positive, b >= 94, and even, and

we already know b<100,
so there are
only

three values to check.
b=94
-> c<100, so doesn't work.
b=96,
c=188, a=1 (any higher a and c < 100)
b=98,
c=285, a=1 (any higher a and c < 100)
** **

Thanks
for the response. Your approach is very neat, and, of course,
generates the correct constants, which would lead to the roots of
(-94, -2) and (-95, -3) for the two sets of answers. I was being
a bit fastidious when I said "by hand" since I was thinking of
eliminating any trial and error at all, but you've pointed out that it
can only be two out of three trials anyway - so hats off to you. The
Puzzlet will still entertain those who prefer to solve these things by
devising an algorithm and writing some code.

Dave.