Ponder This Challenge - February 2026 - Blot-avoiding backgammon strategy
- Ponder This
This riddle was suggested by Asaf Zimmerman - thanks, Asaf!
is called a Pythagorean triple if are positive integers and .
For example, and are all Pythagorean triples.
We know that there's no triple such that , because is irrational, but we can try to get close to that. The triple approximates , but not very well, because , so , but not .
Your goal Find a Pythagorean triple of numbers with 100 digits or less, such that .
Give your solution in the format
A = 24
B = 7
C = 25
A bonus "*" will be given for finding a Pythagorean triple of numbers with 100 digits or less, such that . Give your solution in the same format as above.
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June 2024 Solution:
A solution for both parts is
A = 15780182746062370108602577733859373821827076136375073529796834518989107000489444222482943168070296
B = 5022988173858530362046589594815862687165616836922409678904106986969714780309239151669666891661530
C = 16560331448791934658922466325941893817186620274329261880232738682297512949234184663822339667675754
A generally useful approach for finding rational approximations of numbers is using continued fractions. Given the infinite continued fraction form of a number, one can obtain rational approximations from its convergents - the rational numbers obtained by taking a finite prefix of the continued fraction and computing it. The approximations obtained this way are the best approximations possible for fractions with denominator which is smaller or equal to the one in the approximation.
However, using the continued fraction approach on would not yield approximations that are also a part of a Pythagorean triple. To ensure we have such a triple, we use Euclid's method (from "Elements" book X, proposition 29): given two integers, we can generate a triple by
Fix . We are interested in finding the value of such that . This leads to the simple quadratic equation
With the positive solution with . Thus, a pair such that will give a perfect approximation to , but since we need to be integers and is irrational, we have to settle for an approximation; using the continued fraction method for gives good approximations, solving the riddle.