A238947 Numbers n such that n^2 + 1 = p*q, p < q primes and q-p is a power of 2.

8, 100, 3524, 5084, 36680, 77980, 21474824, 134201344, 148647496, 300741464, 73851531256, 153122539756, 778318386944, 6669171349484, 16526971109344, 596403262068016, 9376599920124524, 26698166963373164, 140144514160214876, 1613032378604451500

Offset

1,1

Comments

Note that if n^2+1 = p*q, then p+q cannot be a power of 2. Proof by contradiction: There are two cases: p an odd prime and p=2. Case 1: suppose p and q are odd primes and q = 2^m-p. Then p(2^m-p) = n^2+1 for some even n. Rearranging terms, we obtain p*2^m-1 = p^2+n^2. Looking at this equation modulo 4, we obtain -1 = 1, a contradiction. Case 2: Let p=2. Then we obtain 2^(m+1)-n^2 = 5, which has no solutions in integers.

Links

Table of n, a(n) for n=1..20.

Giovanni Resta, Some terms greater than a(20)

Example

8^2+1 = 65 = 5*13 and 13-5 = 2^3;
100^2+1 = 10001 = 73*137 and 137-73 = 2^6;
3524^2+1 = 12418577 = 3049*4073 and 4073-3049 = 2^10.

Maple

with(numtheory):for a from 1 to 200000 do:p:=ithprime(a):for i from 1 to 50 do:q:=p+2^i:if type(q, prime)=true then x:=sqrt(p*q-1):if x=floor(x) then print(x):else fi:fi:od:od:

Mathematica

Select[Range[10^6], !PrimeQ[#^2+1]&&Plus@@Last/@FactorInteger[#^2+1]==2&&PrimeNu[#^2+1]==2&&IntegerQ[Log[2, FactorInteger[#^2+1][[2]][[1]]-FactorInteger[#^2+1][[1]][[1]]]]&

Prog

(PARI) isok(n) = (bigomega(n^2+1) == 2) && (f = factor(n^2+1)) && ((f[2, 1] - f[1, 1])== 2^(valuation(f[2, 1] - f[1, 1], 2))); \\ Michel Marcus, Mar 07 2014

Crossrefs

Cf. A134406, A134407, A002522.

Subsequence of A085722.

Sequence in context: A060570 A215875 A317598 * A291536 A229452 A199816

Adjacent sequences: A238944 A238945 A238946 * A238948 A238949 A238950

Keyword

nonn,hard

Author

Michel Lagneau, Mar 07 2014

Extensions

a(7)-a(20) from Giovanni Resta, Mar 07 2014

Status

approved