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A284477
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Pairs of integers (x, y), such that x^2 + 1 and y^2 + 1, 1 < y < x, have the same distinct prime factors.
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1
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7, 3, 18, 8, 117, 43, 239, 5, 378, 132, 843, 377, 2207, 987, 2943, 73, 4443, 53, 4662, 1568, 6072, 5118, 8307, 743, 8708, 2112, 9872, 2738, 31561, 4929, 103682, 46368, 271443, 121393, 853932, 76378, 1021693, 91383, 3539232, 41218, 3699356, 473654
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OFFSET
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1,1
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COMMENTS
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The sequence appears to thin out quite abruptly; however, by solving the Diophantine equation x^2 + 1 = p (y^2 + 1) for a suitable prime p and selecting the solutions (x, y) for which p divides y^2 + 1, it is easy to generate larger pairs, such as (423222288438379883442890018716361, 66096216900526495715353522199871). - Giovanni Resta, Mar 27 2017
A very interesting property: the sequence contains a subsequence of pairs (Lucas numbers L(i), Fibonacci numbers F(i)) for i = 4, 6, 14, 16, 24, 36, ... These pairs are (L(4), F(4)), (L(6), F(6)), (L(14), F(14)), (L(16), F(16)), (L(24), F(24)), (L(26), F(26)), (L(34), F(34)), ... = (7, 3), (18, 8), (843, 377), (2207, 987), (103682, 46368), (271443, 121393), (12752043, 5702887), ... It seems that {i} = A090773(n) (numbers that are congruent to {4, 6} mod 10. - Michel Lagneau, Mar 28 2017
This is because L(i)^2+1 = 5*(F(i)^2+1) for even i, and 5 | F(i)^2 + 1 for i== 3,4,6,7 (mod 10). In fact (L(i), F(i)) for i in A090773 are the solutions of the generalized Pell equation x^2 + 1 = 5 (y^2 + 1) for which 5 | y^2 + 1. - Robert Israel, Apr 10 2017
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LINKS
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EXAMPLE
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The pair (843, 377) is in the sequence because the prime factors of 843^2 + 1 and 377^2 + 1 are 2, 5, 61 and 233.
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MAPLE
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A:= NULL:
for x from 2 to 10^5 do
P:= numtheory:-factorset(x^2+1);
if not assigned(R[P]) then R[P]:= x
else A:= A, op(map(t -> (x, t), [R[P]]));
R[P]:= R[P], x
fi
od:
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MATHEMATICA
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d[n_] := First /@ FactorInteger[n]; Flatten@ Reap[ Do[ dx = d[x^2+1]; Do[ If[ dx == d[y^2+1], Sow[{x, y}]], {y, x-1}], {x, 1, 10^4}]][[2, 1]]
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PROG
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(PARI) upto(n) = {my(l = List(), res=List()); for(i=1, n, f = factor(i^2+1)[, 1]; listput(l, [f, i])); listsort(l); for(i=1, n-1, if(l[i][1]==l[i+1][1], listput(res, [l[i+1][2], l[i][2]]))); listsort(res); res} \\ David A. Corneth, Mar 28 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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