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Triangular primes: or, the dangers of non-rigorous proofs

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This blog post has moved to https://matthewvaneerde.wordpress.com/2008/06/15/triangular-primes-or-the-dangers-of-non-rigorous-proofs/

Comments

  • Anonymous
    June 15, 2008
    3 is an obvious prime triangular number. You found >2 factors but didn't show that they were all different.

  • Anonymous
    June 15, 2008
    Joe:  the different factors is only the case with tn = 1, from what I see. There are two possible sets of factors for a given tn, so find the n where those factors are equal: n = (n + 1) / 2 2 n = n + 1 n = 1 tn = 1 * (1 + 1) / 2 = 1 Second case: n / 2 = n + 1 n = 2 n + 2 n = -2 tn = -2 / 2 * (-2 + 1) = 1 For tn = 3, n = 2.  My suspicion is that there are no such primes greater than 3 to be found, but I don't know that my proof is correct: To be prime the only factors can be 1 and tn.  Therefore at least one of the following has to be true: tn = n tn = n + 1 tn = n / 2 tn = (n + 1) / 2 1: n * (n + 1) / 2 = n (n + 1) / 2 = 1 n + 1 = 2 n = 1 2: n * (n + 1) / 2 = n + 1 n / 2 = 1 n = 2 3: n * (n + 1) / 2 = n / 2 n + 1 = 1 n = 0 4: n * (n + 1) / 2 = (n + 1) / 2 n = 1 Therefore tn primes should be limited to t0, t1, t2, or: tp = { 0, 1, 3 }

  • Anonymous
    June 15, 2008
    @Kfarmer, There is only one prime triangular number, 3. 0 and 1 don't count as they are zeros and unities respectively.

  • Anonymous
    June 17, 2008
    That's you're interpretation (and, admittedly, Dr Math's interpretation), based on how precisely you define primes.   Frankly, I have yet to see One be proven non-prime by any other argument than "we define it as non-prime because we don't want to consider { 1^n } to be a factorization" which is fairly artificial.  I freely admit that I only look for such arguments once or twice a year, and then only briefly.  I have better empires to run. I will grant that Zero is hard to support. Either way, the proof seems to still hold.  The set of tn primes was still limited to that set: whether you consider the set to be a superset of the final answer or not is your problem.

  • Anonymous
    June 17, 2008
    (and before I get flamed for the uniqueness requirements for arithmetic, everybody should remind themselves that math really is just an offshoot of philosophy, which historically has been rife with disagreements) (and, seriously, are you really that bored?)

  • Anonymous
    June 18, 2008
    { t0 = 0, t1 = 1, t2 = 3 } is indeed the set of triangular numbers where the proof breaks down. For t0, the "four distinct factors" are 1, 0, 1, and 0; For t1, they are 1, 1, 1, and 1; For t2, they are 1, 1, 3, and 3. For n >= 3, the "four distinct factors" can be shown to /actually be distinct/ (actually, for primality, it suffices to show that the two internal factors are both at least 2) so the proof is correct from t3 on. The definition of primality I would like to use is "having exactly two factors." 1 has a single factor, and is thus /even better than prime/ -- it's the multiplicative identity. 0 has infinitely many factors but is in turn a factor only of itself, so it's another kind of thing altogether.

  • Anonymous
    October 05, 2008
    Your proof is not "non-rigorous", it simply contains a mistake. In the case where n is even, you fail to observe that the four factors may not be distinct. It is easy to see that the only case in which they aren't distinct is when n=2, and in this case t2=3 is prime. So you have proved that the only prime triangular number is t2. This is not a "non-result", you just have to make the one exception to your hypothesis. So the Huxley quote is completely inappropriate here.