Show that the equation:
987a + 789b + 12321c = (a + b + c)2
has infinitely many integer solutions.
*** Do not include solutions where a+b+c=0.
We only need one infinite series of solutions from potentially many.
Start by calculating 3 small solutions where (a+b)=(a+b)^2:
a b 987a 789b (a+b) sqrt(a+b)^2
1 137 -170 135219 -134130 1089 33
2 1190 -1421 1174530 -1121169 53361 231
3 2639 -3068 2604693 -2420652 184041 421
By inspection, column sqrt(a+b)^2 increases linearly in integer steps without limit
Therefore column (a+b) also increases quadratically in integer steps without limit.
Accordingly there are infinitely many integer solutions.
(a+b) is always positive, and though c is always 0, a+b+c never is.
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Posted by broll
on 2024-07-18 03:10:39 |
Possible Solution