Question 262758
FRom the given, we know that
 {{{a^2 +b^2 = c^2}}}
we also now that c= a+b - 40
By substitution, we get
{{{a^2 + b^2 = (a+b-40)^2}}}
expanding the right side, we get
{{{a^2 + b^2 = a^2 + b^2 + 1600 +2ab -80a - 80b}}}
subtracting a^2 and b^2 from both sides, we get
{{{0 = 1600 + 2ab - 80a - 80b}}}
or
{{{80a + 80b - 2ab = 1600}}}
we can solve for a in terms of b. factor out 2a to get
{{{2a(40 -b) + 80b = 1600}}}
subtract 80b to get
{{{(2a)(40-b) = 1600 - 80b}}}
divide by 2(40-b) to get
{{{a = (80(20-b))/(2(40-b))}}}
which simplifies to
{{{a = (40*(20-b)) / (40-b)}}}
At this point we know that b <= 20 OR b > 40.
Since we want integers for a and b, here are a few options in (a,b) form:
(8,15) or (15,8)
If we use
(8,15), then c = 17 and all 3 share no common factors. But this doesn't satisfy our restrictions.
Our only new options for (a,b,c) are:
(67, 72, 97)