correction, again, > 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. No, it is abc+c^2-c = ab for prime p leading to the same argument as R. Guy's. ralf