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Old 2020-09-27, 08:10   #7
fivemack
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Feb 2006
Cambridge, England

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Default An oddity of this form

It's absolutely obvious in retrospect, but, looking at the results of a match-up-large-primes run, if we have t^2+1=pq then (t+kq)^2+1 will also be divisible by q; and we may occasionally be lucky enough for (t+kq)^2+1 to have only small prime factors other than q.

eg t=18514 has t^2+1=29*11819593 and (t+11819593)^2+1/11819593 is divisible only by 2,5,13,17,29,37, so we can use 18514 and 11838107 instead of 29.

Otherwise the large prime approach isn't all that fertile: p=137593 is the largest to appear twice as the largest factor of an otherwise-17-smooth n^2+1 for n<10^7 (n=1173 and 3853777) ; p=68761153 (n=18542 and 343824307) for 37-smooth and n<10^9

Last fiddled with by fivemack on 2020-09-28 at 19:57
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