Brain teaser
I expected either likes or refutations of the logic. So far neither?
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Can you prove that “With large y there can be no grouping of the factors on the left side into two values that differ by only 2.”?
Consider the figure below that graphs your formulation of the problem: 1729*y^2 = x^2 – 1, solved for y: y = sqrt((x^2-1)/1729). Note the low slope. As x increases a lot, y increases only a little.
The next figure zooms in around y = 4. At x = 166, y is almost, but not quite 4. There are many such points as the curve makes its way to infinity, so we really need an extremely high precision calculator to differentiate these near misses from integers. It’s easy to see how 15 significant digits would not be enough.
Since 1729 is irrational, it seems to me that x^2 -1 must include an odd power of 1729 to cancel that number from the denominator. So we’re looking to solve something like x^2 – 1 = a*(1729^(2n+1)), where a is a perfect square and n>=0.
Just a thought on infinity and such. The astrobiologists tell us that with billions of stars, it’s almost certain that life exists somewhere else in the universe. Here we have an infinite number of integers, so it should be almost certain that one of them fits the bill here.
But almost certain and certain are miles apart in mathematics.
@flga-2-2
Well, there’s no BS in the dungeon, unless it’s in the food.
@mathteacher said “Can you prove that With large y there can be no grouping of the factors on the left side into two values that differ by only 2 ?”
In more detail:
7*13*19*y*y = x^2 – 1 = (x+1)(x-1) says that a solution must allow us to factor each side of the re-organized equation into two integers that differ by 2. There are several possible groupings of the left side into 2 factors, none of which can differ by 2 for large y, in order to match the right side.
(7*13*19) (y*y ) gives two factor with absolute difference (y*y ) – 1729 > 2 for any y >=42 or for any y <=41
A grouping that Includes any additional factors in the group with (y*y) makes the difference larger for ly >=42, and we have checked all y through 42..
With one y in each group, we have four possible groupings that always have absolute difference more than 2 for all y >=1
(7*13*19*y) (y ) difference 1728 y
(7*13*y) (19*y) difference 72 y
(7*y) (13*19*y) difference 240 y
(7*19*y) (13*y) difference 120 y
Therefore there is no integer solution for y>0.
Is there any flaw in this logic?
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Isn’t that where they keep their pet dragon, Spot.? In his case it would be DS, I suppose.
I don’t see anything obviously wrong, but consider using an old problem-solving technique: solving a simpler problem.
Consider 5*y^2+1 = u^2. If y = 4, we have 5*4^2+1 = 81 and x = 9 and (9-1)(9+1) = 80 as it should.
Next consider 7*z^2 + 1 = v^2. If z = 3, we have 7*3^2+1 = 64 and v = 8. (8-1)(8+1) = 63 as it should.
If we multiply these together, we get 35*12^2 + 1 =5041 = x^2 so that x = 71
Perhaps the way to approach the problem is to solve 19*y^2 + 1 = x^2, 13*y^2 + 1 = x^2, and 7*y^2+1 = x^2 and multiply the three results together.
We already have the 7 piece, so it’s the 13 and 19 pieces that we need.
Note also that the easier problem identifies a right triangle with sides 12*sqrt(35), 1, and 71. It’s not a Pythagorean triple, though, so I was wrong about that.
OK, so 13*180^2+1 = 421,201 and sqrt(421,201) = 649. 650*648 = 421,200.
That leaves 19 to be done.
Is there any flaw in this logic?
The flaw was an implicit assumption that y would not be broken up into its factors to be regrouped differently.
In the example of 5 y^2 +1 we get y= 4. My analysis given above would not break apart one of the 4’s into 2’s and so would miss the grouping
(5*2)(4*2) = (8)(10) with 10 – 8 = 2
So I was wrong.
.So your approach may lead to a solution of the original problem, but won’t guarantee it is the solution with smallest y ?
.It may not even lead to a solution. The factor 7 has at least 2 solutions, 3 and 48. Neither will combine with either of the other two factors to produce a perfect square.
19*39^2+1 = 170^2 and 13*180^2+1 = 649^2, but 247*7020^2+1 is not a perfect square.
The idea was to narrow the search area, guessing that the products of the smallest individual solutions would form the smallest solution to the problem. But the product notion appears not to be generally true.
The truth is out there!
Notice all the surveyors dropped out a while ago. Easier to calculate an original corner “off” by 0.07′. Of course, the OP is a surveyor, so there is that.
Yep. Surveyors are smart and don’t waste time on nonproductive pursuits. We old teachers mess with stuff like this because it sometimes leads to a nice enrichment piece.
On the other hand, I don’t teach anymore, so there’s just no explanation.
That leaves 19 to be done.
19* 39^2 + 1 = 28900 and sqrt(28900) = 170
.That’s what I got, but the products don’t work as the did for the example. Maybe there are others that work for all three factors.
@spmpls Maybe we’re just sitting quietly over here in the corner working on the solution. I’ve written a short loop routine using VBA in Excel that is testing all the possibilities. It won’t be a small number that’s for certain.
I’m half way to a billion with no solution yet.
I’m half way to a billion with no solution yet.
Haven’t you exceeded the precision of Excel with the squares when y > 10 million or so? Or did you program a way to keep integer arithmetic going?
.Well, I have an answer by applying the principle of cheating. Enter 1729 At this website:
http://www.jakebakermaths.org.uk/maths/jshtmlpellsolverbigintegerv10.html
I don’t know where Dave got the problem, but here’s one documented source:
https://interestingengineering.com/can-you-solve-this-prison-inmates-viral-math-riddle
It looks like Excel is not up to the task of even checking the answer.
@bill93 I made two mistakes. I had determined that the 3, 180, 39 combo didn’t work so I wrote a loop to find Y’s larger than 180 to test in the factoring method. Then I decided to change the loop to run the original equation rather than just one factor. I goofed and forgot to swap 1729 for 13 on the right side of my equality so it was never going to return a result. My second mistake was misreading my counter (reads in percent complete) and thinking it was at half a billion when it was really at 5 million
Here’s the logic I’m using: If Int(Sqr((1729*y^2)+1)) = Sqr((1729*y^2)+1) then it outputs y. Each time it finds a match it will output to a new cell. I don’t have a great grasp of the limits of Excel’s binary based math limits but I could see a situation where as the numbers get larger I could get false positives due to lack of precision but it should still be a relatively small number of candidates to put through a better calculator.
I ran one for over 18 hours with the messed up formula and an upper limit on the loop of 1 trillion but the percent counter quit updating when the screen went to sleep so I don’t know how far it got. I ran a second one for 7 hours (still with bad formula) and was approaching 40 million when I realized I had a problem and started over.
I’m running this on a 2nd or 3rd gen I-7 processor with a 6 gb graphics card. Running two instances of Excel with this short loop is giving me sustained loads of 60-70% on the CPU and 50-60% on the GPU. This is more load than I can generate with multiple instances each of AutoCAD, QGIS, Chrome, and Netfiex running simultaneously.
It looks like Excel is not up to the task of even checking the answer.
The answer is close to Excel’s limit before it starts losing least digits, before you even think about squaring.
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