Diophantine equation: abcd(1/a + 1/b + 1/c + 1/d)^2 = (a+b+c+d)^2

Determine the nonzero integers   a, \: b, \: c, \: d   such that

a \,b \,c \,d \,(1/a \: + \: 1/b + \: 1/c \: + \: 1/d)^2 \; = \; (a+b+c+d)^2

and such that   a^2 + b^2 + c^2 + d^2   is prime




About benvitalis

math grad - Interest: Number theory
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3 Responses to Diophantine equation: abcd(1/a + 1/b + 1/c + 1/d)^2 = (a+b+c+d)^2

  1. This is definitely going to take some processing power/time. My program currently checks if z=a*a+b*b+c*c+d*d is prime, against the first 100k primes. First output was 2, 3, 640, 960 resulting in 1331213, which is greater than the 100,000th prime, but is not a prime. Was going to run any output against a prime checker, but I might as well just gather them and create a larger list later and just run as is. I also changed my m, max for a, b, c, d from 1k to 10k. There were 178 results for m=100, but none prime.

    1, 4, 284, 1136 ~ 1371169 is my first output for m=10k, but it is not prime…. And there is a ton of output… lol great! I think this list is going to be massive. I should create a prime checker? Technically, the largest output would be 10 quintillion which is not prime, but from 1-10qt there are 279,238,341,033,925 primes… With a=1,b=4, m=10k, there are 289 results alone, with z= 1371169 to 105910289 ha….

    I’ll get back to this in a long time.

  2. Whoops, I guess Z would actually be less than 400,000,000. That helps, but I wrote parts of it anyway to help out. ^_^, but still if there were 1000 checks per second, it still would take 318k years to check with m at 10k

  3. paul says:

    As already pointed out by Daniel this will require a lot of computation time, however, There are a few things that will speed up the search. For the reciprocal part of the equation to be an integer the values of 3 of the terms need to be even and the last one odd, this alone gives a 16 x speed increase. But even with that in place results are not forthcoming so in these instances I like to see some output from my program instead of nothing for hours on end to which the following is output with the difference between the two terms is <= 10. The format is {a, b, c, d, left and right side of equation, the prime part and the difference} there is a near miss with a difference of 1 with {a, b, c, d} = 48,68,72,51. a<b<c<400, d<400



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