Ponder This Challenge - February 2026 - Blot-avoiding backgammon strategy
- Ponder This
Ponder This Challenge:
This month's puzzle comes from S V Nagaraj.
Find four distinct integers a,b,c,d such that
100 <= a,b,c,d <= 12000 and the sum of "all" the four of them
and the sum of "every" two of them are perfect squares.
(That is, a+b, a+c, a+d, b+c, b+d, c+d, and a+b+c+d are all
squares of integers.)
We will post the names of those who submit a correct, original solution! If you don't want your name posted then please include such a statement in your submission!
We invite visitors to our website to submit an elegant solution. Send your submission to the ponder@il.ibm.com.
If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com
The solution is 386, 2114, 3970, 10430. These numbers satisfy the conditions below.
386+2114=2500=50^2
386+3970=4356=66^2
386+10430=10816=104^2
2114+3970=6084=78^2
2114+10430=12544=112^2
3970+10430=14400=120^2
386+2114+3970+10430=16900=130^2
Sketch of a non-brute-force solution:
The sum of all four (a+b+c+d) is the square of the right-hand side of a Pythagorean triple in three different ways, so it has to be divisible by a handful of prime factors, each of which is 1 mod 4 (or else the prime 2). Since it cannot exceed sqrt(4*12000) < 220, there are not that many candidates. The right one turns out to be 130=2*5*13.
If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com