@mathteacher I goofed by overlooking that the?ÿanswer is a perfect square, not a prime number ??? .
@mathteacher?ÿ I tried that calculator and it did give an integer answer:
sqrt((10766518963200)^2??1729+1) = 447685271124001
However, substituting +1 with -100000000000000 gives the same answer:
sqrt((10766518963200)^2??1729??100000000000000) = 447685271124001
and
sqrt((10766518963200)^2??1729??1000000000000001) = 447685271124000
So that calculator obviously doesn't have enough significant figures to prove that y=10766518963200.?ÿ Must be the y?ý operation exceeds its range as:
sqr(13!) is reported as = 3.87757880436326e+19 which has been truncated.
?ÿ
@mike-marks; I get for 13! 6227020800 and the square root as 78911.474450804681438 rounded
in the last place. How do you get sqr(13!) as 3.8857880+e+19?
Do I have something wrong or you??
Thanks,?ÿ JOHN NOLTON
Where did the factorial part of the brain teaser come from??ÿ It isn't in the original post.
Factorials and powers of 2 are staples of prime number hunters. A lot of cryptography is based on numbers that are not prime but are next to impossible to factor. Not many people are good at finding those factors, so very secure encryption is possible.
@john-nolton?ÿ I have something wrong.?ÿ Ignore my last statement in my above post concerning 13!= (some big number); I suspect my cat jumped on the keyboard.?ÿ Yes the sqrt(13!) = 78911.474450804681438 but that's not what we are trying to solve which is:
What is the smallest positive integer, y, that makes 1729y?ý+1 equal a perfect square??ÿ?ÿI.e., the sqrt(1729y?ý+1)?ÿis an integer which means with a correct y value it's a perfect square.
Bugger reported the solution as y = 1729 * 13! which =?ÿ 10766518963200.
sqr(y) = 115,917,930,584,945,202,954,240,000
sqr(y)*1729 = 200422101981370255907880960000
sqr(y)*1729 + 1 = 200422101981370255907880960001
sqrt(sqr(y)*1729 + 1) = 447685271124001 according to the "hi-res" calculator.
But as I showed above even the "hi-res" calculator has lost so many significant digits it's not proving Bugger's value for y is the solution.
Kind of interesting to note that the problem can be cast as a right triangle with legs y*sqrt(1729) and 1 and hypotenuse sqrt(1729*y*y +1).
One acute angle is tiny and the other is almost a right angle. The problem is a really strange Pythagorean triple.
@bugg All whole numbers are?ÿintegers, so since 0 is a whole number, 0 is also an?ÿinteger.?ÿ Therefore y=0 is the correct answer.
1729 * sqr(0) +1 = 1 which is a perfect square.?ÿ 0 is the smallest integer so it is the correct answer.?ÿ I'm concerned a negative integer may provide a "smaller" value for y which y is less than -1 but I'm not holding my breath.
If y=0 isnt ok
Zero is not OK as it is not a positive integer.
I pray Lindell reveals the answer soon as I'm spending way too much time puzzling this out.
You have to use the full precision calculator and set the number of decimal places you want in the little box below the "history" box. Something like 50 is pretty good, but a 1 in the 51st place would rule out an integer.
You have to use the full precision calculator and set the number of decimal places you want in the little box below the "history" box. Something like 50 is pretty good, but a 1 in the 51st place would rule out an integer.
I used the scientific calculator in my calculations above which has no stated sig fig accuracy but have figured out how modify the solution to use the basic calculator @ 800 decimals (which I think they mean significant figures):
?ÿ((13!??1729)^2 *1729 + 1)^0.5 =
447685271124000.96907868197395816017025777903449911382572443851297095521666106751218437070058205777176149485201524466825229216867944449799510176171751956731064601672662241080327806696590921976971197214465447266747395546960051246161196755712797607325717505004027961115897270959908331962301000865788505978263895235013655475452773437898415025265382652603875085352746887042224696415170772415880569727832567180285985450294452769529805305340746884928358473006284839236697163098599557817962626393616214637699205784295230129164700410152587466914913466882859114837276585034731887516618948836054204453772992916564204834136426483826868707056871077228262692709910482584502896120326308186619355364214623454633409532236237017859938738391813417723184914592837790137677578307399802651159273193308595975000909304985881?ÿ
So bugg's solution that y= 13!??1729 is sorta buggy unless some strange rounding is occurring in the intermediate steps of the above calculation.?ÿ It doesn't result in a perfect square integer, though it's pretty darn close.
I think it's something like 6 or 7 decimal places of Pi required to navigate across the solar system, but that may not be enough to identify an integer. I was never a good number theorist, but this one may require that.
1729 is the Ramanujan taxi-cab number. I saw a YouTube video on it.?ÿ
1729 = 12^3 + 1^3 = 10^3 + 9^3
"it is a very interesting number; it is the smallest number expressible as the sum of two [positive] cubes in two different ways."
But the important 'thing' about 1729 is that:
1729^17 + 1729^29 - 1 is a Pythagorean prime: (4n+1)?ÿ
And a perfect square has even powers of its primes. For example:
5336100 = 2310^2 = ( 2?ÿ * 3 * 5 * 7 * 11 ) ^2
The primes of 1729 are 7, 13, 19: 7 * 13 * 19 = 1729
(Interesting difference of 6 here: 13-7=6 and 19-13=6)
?ÿ
Zero is not positive, so is ruled out by the problem statement.
Restate the problem as
7*13*19*y*y = x^2 - 1 = (x+1)(x-1)
We are looking for two integers on the right side differing by only 2, which are also factors of the left side.
We have found numerically that no small y works.
With large y there can be no grouping of the factors on the left side into two values that differ by only 2.
Therefore there is no positive integer solution.
This thread is great, it??s like Herman and Grandpa in the dungeon arguing over ingredients for a magic pill. ?????ÿ
@flga-2-2
?ÿ
If you like this, then oeis will eat up the rest of your day.
As will numberphile 1 and 2 on YouTube.?ÿ
?ÿ
Get youtube premium so you dont have to sit through commercials.
?ÿ
Get ready to blow the rest of 2021...
