My brother asked me for some help with a novel he’s writing, he wants a character in his book to solve a math riddle. I know what I want the answer to be, I just don’t know if the riddle I came up with is too abstruse and needs improvement. Here’s what I’ve got:
“Pythagoras said ‘Number rules the universe’
So here’s a number: 1,940,449
If you know the half of it (more or less)
and get to the root of the matter,
Then you may hit a triple
and calling us will be as easy as A-B-C
(you’ll have won at the end).”
Okay, if you have no idea wtf I’m talking about:
Summary
The answer is a ten-digit telephone number. The riddle is intended to decode into the Pythaorean triple 696-697-985, with 1 as the last digit of the number. By intention not a valid area code.
I got as far as 985, and that was it. Even if I knew what a Pythagorean triple was before now (other than 3,4,5), I would not have recognized that 985 was one.
(Note that 985 is apparently the hypotenuse of at least four Pythagorean triples, and the leg of four more. The one you are trying to get to is the only primitive Pythagorean triple in which the sides are consecutive integers, though. Maybe that could be another clue?)
And I thought calling ABC had something to do with dialing 222 on a standard telephone keypad.
Sorry, I would never have gotten this with only these clues.
Unless the book is intended for math puzzlers, not one in 10K readers will solve that. Now they may well accept that some character in the book can solve it, and just take as given the phone number the character works out.
But if your author has a goal of having readers solve this themselves, ain’t gonna happen wo invoking AI. And probably not then.
Is the character who solves this supposed to be a math genius or something? I think we need to know a little background on the character to tell if this is a reasonable riddle for them to solve. But I’d say that even if the character is Ramanujan, it’s pretty abstruse.
I somehow typo’d the OP’s base number in my calculations in my earlier post. So I was dead-ended from the start, leaving me to think it was a rather harder problem than it really is.
Ahem … Upon Further Review …
But once you trivially derive 985, what next? There are lots of kinds of math triples that are not Pythagorean. I think the riddle might benefit from some manner of reference to a triangle at that point. e.g. “… root of the matter, then you could rightly triangulate. Then calling us will be …”
But even if we correctly guess that the riddle as written wants a specifically Pythagorean triple, we have this little issue:
Under the current NANP dialing rules, any of the last 3 are valid North American phone numbers format-wise. At least they are once you tack on the trailing won one.
And there is no specific requirement within Pythagorean triples that the three numbers be specified in sorted order. So each one of those three NANP-compatible triples can be reordered in any one of 6 different orders. Now you have 18 possible phone numbers. And the first one, starting with 140, can also be reordered into 4 sequences that are NANP-valid. Now there are 22 equally valid answers to this riddle.
Even if you assert that the third entry C must be the hypotenuse, then A & B can still appear in either order. Which leads to 7 distinct correct answers.
Bottom line: Needs work. It’s close, but not yet show-ready.
(696+697)² = 1,940,449. That can be split into 970,225 the square root of which is 985, and 970,224 which is 2×696×697. The difference by 1 is pivotal, but I couldn’t think of a way to get across how.
The only Pythagorean triple consisting of 3 consecutive integers is (3,4,5).
The proof is pretty simple. Let n be the middle number, so (n-1)^2 + n^2 = (n+1)^2.
Then (n^2-2n+1)+n^2 = n^2+2n+1
Simplify: n^2-4n = 0 n(n-4) = 0
So the only two solutions are either n=0 or n=4.
The first case yields the triple (-1,0,1) which isn’t Pythagorean because of the negative number, and the second case yields (3,4,5).
A similar argument shows that for consecutive values A,B,C, if A^2+B^2 = C^2+1, the only solution is B = 2+\sqrt{5} and if A^2+B^2 = C^2-1, the only solution is B = 2+\sqrt{3}, so there are no such integer triples.
Here is the rule for generating primitive PTs (no common divisors). Take 2 numbers m and n, no common divisors and one even (and the other odd). Assume m > n. Then 2mn, m^2-n^2, m^2+n^2 is a primitive PT and that is the only way they can be generated.
So if you want 985 to be the hypotenuse of a PT, it has to be the sum of two squares. 985 = 5 x 197. 5 = 1^2+2^2 and 197=14^2+1^2 and a simple formula implies that
985=(2 x 14 + 1 x 1)*2 + (2 x 1- 14 x1)^2 = 29^2 + !2^2.
So take m = 29 and n=12. Then m^-n^2=697 and 2mn = 696, so the PT is 696, 797, 985.
Another solution using a different way of multiplying sums of two squares is 473, 864, 985. It is also possible to make 985 the difference of two squares and find solutions in which 985 is a leg.
