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

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

{4,6,12,9,968,961,277,7}
{4,8,12,3,722,729,233,7}
{10,14,30,21,5618,5625,1637,7}
{12,18,20,15,4232,4225,1093,7}
{18,20,40,45,15125,15129,4349,4}
{20,56,60,21,24642,24649,7577,7}
{24,48,68,17,24642,24649,7793,7}
{30,42,50,35,24642,24649,6389,7}
{32,44,48,33,24642,24649,6353,7}
{40,50,56,35,32768,32761,8461,7}
{40,54,60,45,39605,39601,10141,4}
{42,48,56,63,43687,43681,11173,6}
{48,50,80,75,64000,64009,16829,9}
{48,68,72,51,57122,57121,14713,1}
{52,78,90,45,70227,70225,18913,2}
{60,76,120,95,123210,123201,32801,9}
{66,70,110,105,123210,123201,32381,9}
{76,108,114,81,143648,143641,36997,7}
{80,110,112,77,143648,143641,36973,7}
{88,124,132,93,190962,190969,49193,7}
{90,126,130,91,190962,190969,49157,7}
{100,112,120,105,190962,190969,47969,7}
{110,132,140,105,237160,237169,60149,9}
{116,328,348,123,837218,837225,257273,7}
{168,170,360,357,1113032,1113025,314173,7}
{168,194,336,291,978123,978121,263437,2}
{168,200,210,175,567000,567009,142949,9}
{170,190,228,255,710645,710649,182009,4}
{174,246,290,205,837218,837225,216917,7}
{216,306,312,221,1113032,1113025,286477,7}
{220,308,310,217,1113032,1113025,286453,7}
{234,260,270,225,978123,978121,245881,2}
{240,272,288,255,1113032,1113025,279553,7}

Paul.