You may remember that in yesterday’s post titled “Monday Miscellany” I included one of the problems I had solvced on the mathematical website Brilliant – this post presents the solution and also clears up a side issue raised on that website by disgruntled folk who had got it wrong.
Here is the problem again from yesterday:
The answer is True. The formula for (x+1)^2 is X^2 + 2X + 1, and every odd number greater than 1 could serve as the 2X+1 part of that equation.
THE SIDE ISSUE
Some people on Brilliant cavilled at this because there are some Primitive Pythagorean Triples whose smallest term is even (8,15, 17, 20,21,29 and 65,72,97 were all mentioned, although none of the complainers mentioned 12,35,37, 60,91,109 or 696,697,985). The question did not state that the triple of which the odd number is the lowest term was the lowest triple to feature that number, and indeed if one looks carefully at the triangles presented as part of the problem one can see clearly that the odd number is allowed to be in another triple where it is not the lowest term:
Note that the number 5 features twice (ringed in the diagram above, once as the largest term in a triple and once as the smallest).
Thus that 15 features in 8,15,17 does not invalidate the claim of the question since it is the smallest term in 15, 112, 113. All the other odd numbers mentioned in triples of which they are not the smallest member likewise feature in triples in which they are the smallest member, the biggest being 985, 485112, 485113.