Each of a, b, and c is a different base ten digit and n is a positive integer such that:
ab2+c2=ac2+cb2=n
For example:
272+12=212+172=730
Determine the total number of values of the positive integer constant n less than 2023, such that we will have valid values of a,b, and c satisfying the abovementioned condition.
***** No number can contain any leading zero.
So for algebra, lets expand ab, ac and cb into 10a+b, 10a+c and 10c+b.
Then (10a+b)^2 + c^2 = (10a+c)^2 + (10c+b).
Expand and cancel like terms leaves 20ab = 20ac + 100c^2 + 20bc.
Then I'll divide out the common factor 5 and then rearrange a bit to factor in terms of c. That leaves (a-c)*(b-c) = 6c^2.
If c>=3 then the right side is at least 6*3^2=54, be then the left side is at most (9-3)*(8-3)=30. So no solutions for c>=3.
If c=2 then (a-2)*(b-2) = 24. With a and b being at most 9, then a-2 and b-2 are at most 7. So we need factorizations of 24 that are at most 7. There is one choice: 24=4*6.
Then we have two potential solutions with a=6, b=8 and a=8, b=6.
If a=6, b=8, c=2 then 68^2+2^2 = 62^2+28^2 = 4628. This is a solution to the equation but lays outside the 2023 limit.
If a=8, b=6, c=2 then 86^2+2^2 = 82^2+26^2 = 7400. This is also a solution to the equation but again lays outside the 2023 limit.
If c=1 then (a-1)*(b-1) = 6. Both factorizations of 6, 1*6 and 2*3, are possible for this case. This gives potential pairs a and b as a=2,b=7; a=7,b=2; a=3,b=4; and a=4,b=3.
If a=2, b=7, c=1 then 27^2+1^2 = 21^2+17^2 = 730. This is a solution.
If a=7, b=2, c=1 then 72^2+1^2 = 71^2+12^2 = 5184. This is a solution to the equation but lays outside the 2023 limit.
If a=3, b=4, c=1 then 34^2+1^2 = 31^2+14^2 = 1156. This is a solution.
If a=4, b=3, c=1 then 43^2+1^2 = 41^2+13^2 = 1850. This is a solution.
Thus, the total number of values of the positive integer constant n less than 2023 is three (730, 1156, and 1850).
Solution
