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A217277
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Numbers n such that n^2-9 is divisible by consecutive primes beginning with 2.
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3
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5, 9, 15, 21, 27, 33, 51, 57, 87, 93, 123, 147, 213, 297, 333, 483, 753, 1053, 1347, 2307, 2643, 3237, 4287, 6003, 10293, 12477, 14403, 18147, 26247, 58803, 74253, 739203, 1166883, 3801333
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OFFSET
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1,1
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COMMENTS
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Searched to 10^8. If this sequence is finite with 34 terms, then there is only one solution to m! + 9 = n^2: m=6, n=27.
Consider the Diophantine equations m^2 - 9 = k*(n^2 - 9), which lead to the Pell-like equations m^2 - k*n^2 = 9*(-k + 1) (1) where the only primes dividing k are in P = {2,3,5,7,11,13,17,19} and such that for every k there is at least one solution (m,n) that belongs to the sequence.
As an example, if k = 10 = 2*5, the 2 pairs of solutions of m^2 - 10n^2 = -81; (27,9) and (333,1053) belong to the sequence but the other solutions seem to not be P-smooth. If k = 30 = 2*3*5; m^2 - 30n^2 = -261 and (147, 27) is a solution belonging to the sequence, ...
If the infinitely many solutions of the Pell-like equations are never P-smooth, then this sequence is finite and there is a precise answer to the extended Brocard's problem: There are exactly 3 solutions to m!+t^2 = n^2 with t=1; 1 solution with t = 3; ... - Robin Garcia, Oct 01 2012
All terms beyond the first are 3 mod 6, since otherwise n^2 - 9 = 2^k for some k and so n - 3 and n + 3 are both powers of 2. For n > 7 all terms are +-3 mod 30, a consequence of Størmer's theorem. - Charles R Greathouse IV, Oct 01 2012
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LINKS
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EXAMPLE
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5^2 - 9 = 2^4; 9^2 -9 = 2^3*3^2; 27^2 -9 = 2^4*3^2*5 = 6!; 87^2 - 9 = 2^3*3^3*5*7; 333^2 - 9 = 2^5*3^2*5*7*11.
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PROG
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(PARI) is(n)=my(m=n^2-9, t); forprime(p=2, , t=valuation(m, p); if(t, m/=p^t, return(m==1)))
print1(5); forstep(n=9, 1e6, 6, if(is(n), print1(", "n)))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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