Date: Wed, 9 May 2007 16:42:32 -0700 From: Dean Hickerson (dean(AT)math.ucdavis.edu) Proof that a(15) = 0. With a postscript from Tony Noe, May 09 2007 Nick Hobson wrote: > Suppose p and q are consecutive primes. Let > > a = sqrt((p^2 + q^2)/2 - 1). (1) > > It is easy to see that if q = p + 2 then a is an integer. Is the converse > true? > > If we consider the smallest prime q > p + 2 which makes a an integer, we > generate the following sequence, where 0 is used if there is no such q: > > 11, 263, 59, 23, 101, 109, 0, 151, 193, 79, 269, 277, 311, 0, 179, 83, > 83003, 479, 487, 181, 563, 571, 613, 1201, 157, 141509, 739, 773, 479 Don Reble replied: > I get p=19,q=1278886952463697; p=127,q=6858037981; > but I find no q for p=47. There is no q for p=47. For this case, (1) simplifies to the Pell-type equation q^2 - 2 a^2 = -2207. (2) I claim that for every integer solution of this, q is divisible by either 3, 5, 7, or 11, so it can't be prime. Equation (2) describes a hyperbola. Given one point (q,a) on the a>0 branch, we can construct infinitely many others (q', a') on the same branch by writing q' + a' sqrt(2) = (q + a sqrt(2)) (3 + 2 sqrt(2))^n (3) where n is an integer. Every point on the a>0 branch is equivalent in this way to exactly one with -q < a <= q. Combining that inequality with equation (2) implies 33 < sqrt(2207/2) <= a <= sqrt(2207) < 47. Checking the values of a in that range produces just two integer solutions: (q,a) = (45,46) and (q,a) = (-45,46). So every integer solution of (2) with a>0 is given by (q,a) = (45 + 46 sqrt(2)) (3 + 2 sqrt(2))^n (4) or (q,a) = (-45 + 46 sqrt(2)) (3 + 2 sqrt(2))^n (5) for some integer n. Note that (3 + 2 sqrt(2))^12 = 768398401 + 543339720 sqrt(2) = 1 + 1155 (665280 + 470424 sqrt(2)) So if q' + a' sqrt(2) = (q + a sqrt(2)) (3 + 2 sqrt(2))^12 (6) then q' == q (mod 1155). Note that 1155 = 3*5*7*11. So to verify that the q's given by (4) and (5) are always divisible by 3, 5, 7, or 11, it suffices to check 12 consecutive values in each sequence: n q from (4) a from (4) q is divisible by ----------------------------------------------------- 0 45 46 3, 5 1 319 228 11 2 1869 1322 3, 7 3 10895 7704 5 4 63501 44902 3 5 370111 261708 7 6 2157165 1525346 3, 5 7 12572879 8890368 11 8 73280109 51816862 3, 7 9 427107775 302010804 5 10 2489366541 1760247962 3 11 14509091471 10259476968 7 n q from (5) a from (5) q is divisible by ----------------------------------------------------- 0 -45 46 3, 5 1 49 48 7 2 339 242 3 3 1985 1404 5 4 11571 8182 3, 7 5 67441 47688 11 6 393075 277946 3, 5 7 2291009 1619988 7 8 13352979 9441982 3 9 77826865 55031904 5 10 453608211 320749442 3, 7 11 2643822401 1869464748 11 So q is never prime. Dean Hickerson Postscript from Tony Noe: Dean, I agree with your analysis. Don Reble and I have been working on this problem off-list. I have computed 1000 terms and Don has computed even more. I have done all my work with recursive equations instead of the radicals you use. For instance, after the q(0) and q(1) terms are known, the second column of your tables can be computed using the recursion q(n+1) = 6*q(n) - q(n-1) To prove that there is no possible solution for a prime such as 47, only the first 6 terms need to be computed. Using this method, it can be shown that there is no possible solution for the primes 47, 443, 1867, 6883, 13417 and 13907. Tony