6 ms·

So what was the math problem?

by vex 12y ago

sillysaurus3 12y ago
Find integers a, b, c such that a^20 + b^20 = c^20. The solution turned out to be 4110^20 + 4693^20 = 4709^20. You can verify this is correct using any old calculator, for example: https://www.google.com/search?q=4110%5E20+%2B+4693%5E20 https://www.google.com/search?q=4110%5E20+%2B+4693%5E20 https://www.google.com/search?q=4709%5E20 https://www.google.com/search?q=4709%5E20
- bowmessage 12y ago
  Are you sure those were the only restrictions on the problem? Wouldn't [1, 1, 2^(1/20)] also be a solution?
  - sillysaurus3 12y ago
    Whoops, fixed. I forgot to say that a, b, c must be integers. Thanks!
- wbhart 12y ago
  This particular solution only works on April 1st. However 0^20 + 1^20 = 1^20 works any day of the year.
  - TheLoneWolfling 12y ago
    In general, a^20 + 0^20 = ((+ or -)a)^20, or 0^20 + a^20 = ((+ or -)a)^20.
- dil8 12y ago
  Am I missing something, doesn't this contradict Fermat's last theorem > In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation a^n + b^n = c^n for any integer value of n greater than two. https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
  - TheLoneWolfling 12y ago
    You're not missing anything. His numbers do not add up to the same thing. In other words, 4709^20 != (4110^20 + 4693^20). (The difference is ~10^61 or so, whereas the numbers are ~10^73. In other words, they diverge at ~ the 12th digit, whereas many calculators only display 10.)
    - sillysaurus3 12y ago
      Hm, well, part 2 of the question was to find a solution for a^15 + b^15 = c^15, where a, b, c are integers > 0. Google verifies the answer is 434437^15 + 588129^15 = 588544^15: https://www.google.com/search?q=434437%5E15+%2B+588129%5E15+-+588544%5E15 https://www.google.com/search?q=434437%5E15+%2B+588129%5E15+...
      - TheLoneWolfling 12y ago
        Nope. This is not correct either. > >>> 434437* * 15 + 588129* * 15 - 588544 * *15 > -604550152144288043930860169354171954730939671404246170822386878582482 (Edit: how do I display two asterisks in a row? It's supposed to be number(asterisk)(asterisk)number.) This is using Python, which does arbitrary-precision integer arithmetic. Google's calculator probably uses floating-point numbers internally, and hence starts losing precision.
        sillysaurus3 12y ago
        It seems that advancements in technology have made mathematical trolling much more difficult. :) In case anyone is curious, the above "solutions" are called near-misses, since they're almost correct. A clever person came up with an algorithm to generate interesting near-misses for low exponents. See the table on page 15: http://arxiv.org/pdf/math/0005139v1.pdf http://arxiv.org/pdf/math/0005139v1.pdf Related: http://math.stackexchange.com/questions/526330/fermats-last-theorem-near-misses http://math.stackexchange.com/questions/526330/fermats-last-... Have fun!
        12y ago
- TheLoneWolfling 12y ago
  Umm... About that: https://www.google.com/search?q=4709^20+-+%284110^20+%2B+4693^20%29 https://www.google.com/search?q=4709^20+-+%284110^20+%2B+469... Just because they don't disagree on the significant figures you can see doesn't mean there isn't a difference.
- diziet 12y ago
  Andrew Wiles would like to have a word with you regarding this claim!
- IvyMike 12y ago
  Also see this same gag on The Simpsons http://www.npr.org/blogs/krulwich/2014/05/08/310818693/did-homer-simpson-actually-solve-fermat-s-last-theorem-take-a-look http://www.npr.org/blogs/krulwich/2014/05/08/310818693/did-h...
- lordnacho 12y ago
  Wait a minute, is it somehow clear that you're supposed to be looking for a near-miss? If I came across this in my line of inquiry, I would assume that my previous step was wrong, and I'd have dropped the puzzle eventually.
  - sillysaurus3 12y ago
    Nah, I didn't actually do the Cicada 3301. I was just joking around. It seemed unlikely anyone was going to post an interesting math problem, so I decided to have a little fun. https://news.ycombinator.com/item?id=8549204 https://news.ycombinator.com/item?id=8549204
- JoshTheGeek 12y ago
  Fermat's last theorem has been proven; there is no solution, people. In this case, you can't verify this using "any old calculator," as most show only ten digits and these diverge at digit 12.

So what was the math problem?

by vex 12y ago

So what was the math problem?

Find integers a, b, c such that a^20 + b^20 = c^20. The solution turned out to be 4110^20 + 4693^20 = 4709^20. You can verify this is correct using any old calculator, for example: https://www.google.com/search?q=4110%5E20+%2B+4693%5E20 https://www.google.com/search?q=4110%5E20+%2B+4693%5E20 https://www.google.com/search?q=4709%5E20 https://www.google.com/search?q=4709%5E20

bowmessage 12y ago
Are you sure those were the only restrictions on the problem? Wouldn't [1, 1, 2^(1/20)] also be a solution?
- sillysaurus3 12y ago
  Whoops, fixed. I forgot to say that a, b, c must be integers. Thanks!
wbhart 12y ago
This particular solution only works on April 1st. However 0^20 + 1^20 = 1^20 works any day of the year.
- TheLoneWolfling 12y ago
  In general, a^20 + 0^20 = ((+ or -)a)^20, or 0^20 + a^20 = ((+ or -)a)^20.
dil8 12y ago
Am I missing something, doesn't this contradict Fermat's last theorem > In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation a^n + b^n = c^n for any integer value of n greater than two. https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
- TheLoneWolfling 12y ago
  You're not missing anything. His numbers do not add up to the same thing. In other words, 4709^20 != (4110^20 + 4693^20). (The difference is ~10^61 or so, whereas the numbers are ~10^73. In other words, they diverge at ~ the 12th digit, whereas many calculators only display 10.)
  - sillysaurus3 12y ago
    Hm, well, part 2 of the question was to find a solution for a^15 + b^15 = c^15, where a, b, c are integers > 0. Google verifies the answer is 434437^15 + 588129^15 = 588544^15: https://www.google.com/search?q=434437%5E15+%2B+588129%5E15+-+588544%5E15 https://www.google.com/search?q=434437%5E15+%2B+588129%5E15+...
    - TheLoneWolfling 12y ago
      Nope. This is not correct either. > >>> 434437* * 15 + 588129* * 15 - 588544 * *15 > -604550152144288043930860169354171954730939671404246170822386878582482 (Edit: how do I display two asterisks in a row? It's supposed to be number(asterisk)(asterisk)number.) This is using Python, which does arbitrary-precision integer arithmetic. Google's calculator probably uses floating-point numbers internally, and hence starts losing precision.
      - sillysaurus3 12y ago
        It seems that advancements in technology have made mathematical trolling much more difficult. :) In case anyone is curious, the above "solutions" are called near-misses, since they're almost correct. A clever person came up with an algorithm to generate interesting near-misses for low exponents. See the table on page 15: http://arxiv.org/pdf/math/0005139v1.pdf http://arxiv.org/pdf/math/0005139v1.pdf Related: http://math.stackexchange.com/questions/526330/fermats-last-theorem-near-misses http://math.stackexchange.com/questions/526330/fermats-last-... Have fun!
        12y ago
TheLoneWolfling 12y ago
Umm... About that: https://www.google.com/search?q=4709^20+-+%284110^20+%2B+4693^20%29 https://www.google.com/search?q=4709^20+-+%284110^20+%2B+469... Just because they don't disagree on the significant figures you can see doesn't mean there isn't a difference.
diziet 12y ago
Andrew Wiles would like to have a word with you regarding this claim!
IvyMike 12y ago
Also see this same gag on The Simpsons http://www.npr.org/blogs/krulwich/2014/05/08/310818693/did-homer-simpson-actually-solve-fermat-s-last-theorem-take-a-look http://www.npr.org/blogs/krulwich/2014/05/08/310818693/did-h...
lordnacho 12y ago
Wait a minute, is it somehow clear that you're supposed to be looking for a near-miss? If I came across this in my line of inquiry, I would assume that my previous step was wrong, and I'd have dropped the puzzle eventually.
- sillysaurus3 12y ago
  Nah, I didn't actually do the Cicada 3301. I was just joking around. It seemed unlikely anyone was going to post an interesting math problem, so I decided to have a little fun. https://news.ycombinator.com/item?id=8549204 https://news.ycombinator.com/item?id=8549204
JoshTheGeek 12y ago
Fermat's last theorem has been proven; there is no solution, people. In this case, you can't verify this using "any old calculator," as most show only ten digits and these diverge at digit 12.