A187401 Numbers k such that k^2 + 1 = p*q, p and q primes and |p-q| is square.

30, 100, 144, 274, 484, 516, 526, 756, 1046, 1250, 1714, 1806, 1834, 2284, 2440, 2610, 2940, 3524, 3824, 4190, 5084, 5746, 6766, 7486, 9746, 9920, 10310, 13024, 13210, 15396, 16916, 17546, 18726, 19256, 20000, 21194, 23214, 24964, 30370, 30394, 31126, 31496, 35180, 36680, 37816

Offset: 1

Author

Michel Lagneau, Mar 09 2011

Keywords

• nonn

Comments

Note that if k^2+1 = p*q, then p+q cannot be a square. 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 = y^2-p. Note that y must be an even number in order for q to be odd. Then p(y^2-p) = x^2+1 for some even x. Rearranging terms, we obtain p*y^2-1 = p^2+x^2. Looking at this equation modulo 4, we obtain -1 = 1, a contradiction. Case 2: Let p=2. Then we obtain 2y^2-x^2 = 5, which has no solutions in integers. - T. D. Noe, Mar 10 2011

Examples

			20000 is in the sequence because 20000^2+1 = 19801 * 20201 and 20201 - 19801 = 20^2.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..600

Crossrefs

Cf. A134406, A134407, A002522, A005574.

Programs

Maple

with(numtheory):nn:=50000:for i from 1 to nn do: n:=i^2+1:x:=factorset(n):x1:=nops(x):x2:=bigomega(n):if  x1=2 and x2=2 then z:=x[2]-x[1] :w:=sqrt(z):if w= floor(w) then printf(`%d,  `, i):else fi:else fi :od:
# Alternative:
N:= 500: # to get a(1) to a(N)
count:= 0:
for k from 2 by 2 while count < N do
  f:= ifactors(k^2+1)[2];
  if nops(f) = 2 and {f[1,2],f[2,2]}={1} and issqr(abs(f[1,1]-f[2,1])) then
    count:= count+1;
    A[count]:= k;
  fi
od:
seq(A[i],i=1..count); # Robert Israel, Jun 09 2014

Mathematica

okQ[k_] := Module[{ff = FactorInteger[k^2+1]}, Length[ff] == 2 && ff[[All, 2]] == {1, 1} && IntegerQ[Sqrt[ff[[2, 1]] - ff[[1, 1]]]]];
Select[Range[2, 40000, 2], okQ] (* Jean-François Alcover, Jun 25 2019 *)

Sage

A = []
for k in range(2, 2000, 2):
    K = k^2 + 1
    f = prime_divisors(K)
    if len(f) == 2:
        if mul(f) == K:
            if is_square(abs(f[0]-f[1])):
                A.append(k)
print(A) # Peter Luschny, Jun 10 2014