Positive integers a and b are such that (15a+16b) and (16a-15b) are perfect squares. Find the least possible value of the smaller of these two squares.

(In reply to Solution - not sure if least by goFish)

good job goFish!  

I had 481 at the top of my list for one of the squared numbers, but hadn't tried it as a or b.  I suppose that assuming both squares were equal should have been an obvious route.  Your solution falls right out when you do that.  I haven't been able to prove your value is the least value for one of the squares yet, but it sure looks that way at the moment. 

Here's some steps that might help lead towards a proof:

Any solution requires a ≥31b.  Write the two equations
     15a + 16b =x²      and
     16a - 15b = y². 
Introduce the variable c, such that a=31b+c and substitute this for a in both equations.  Determine the sum and the difference between the two equations.  You end up with
     962b=32x²-30y²    and
     c=y²-x²;
Substituting c for y²-x² in the former equation results in:
     x²=15c+481b.
We know that x² must be the smaller of the two squares, and it is obvious that one way to arrive at a smaller value for x², we might want to minimize the value of c.  c cannot be negative, so c=o is an obvious choice, leaving x²=481b.  Obviously, b=481 in this case and this leaves us with your solution, goFish.  What we haven't yet proven (but it shouldn't be too hard, I think) is that this solution provides the lowest value for x².  There may be a lower value for b (requiring a higher value of c) that also meets the requirements and results in a lower value for x².  There may be...but I doubt it.

Good job.

Posted by Mindy Rodriguez on 2005-11-07 21:48:58