# summary Re: square (prime+square)

Ralf Stephan ralf at ark.in-berlin.de
Sat Mar 29 17:21:35 CET 2003

```%N A075555 Smallest prime p such that p+n is a square, 0 if there is no such p.
%C A075555 a(n)=0, at least for n=A047845(i)^2. Proof: If there is (k+a)^2 = k^2 + a(2k+a) then a would have to be 1, and 2k+1 prime, which does not hold for those n (R. Guy). If Schinzel's hypothesis is true, then these n are the only ones so that a(n)=0 (D. Hickerson).

I will express my agreement with Don Reble's yes-koan by giving yet
another proof:
Prove: If 2k+1 is composite (=ab) then there is p prime that k^2+p is (k+c)^2.
Assume p+(ab-1)^2/4=((ab-1)/2+c)^2, then p would have to be
abc+c^2-c-ab/4. For p to be integer, 4 must divide ab which
contradicts oddness of ab.

Regarding M. LeBrun's advice, I would like to see it in a FAQ list,
if it isn't already.

Finally, many thanks to Neil Sloane for the service, I hope he sees
that I cannot thank him all the time, better do it once in a while.

ralf

```