Each of 2N + 1 and 3N + 1 is a perfect square for a positive integer N.

Can 5N + 3 be a prime number?
If not, prove it.
If so, provide an example.

Algebraic sleight of hand only:

I (2n+1)=a^2, (3n+1)=b^2, so

(i) both n=(b^2-a^2) and n=a^2+1/2(a^2-1)
(ii) b^2=a^2+1/2 (a^2-1)
(iii) 3a^2-2b^2 = 1 (doubling, collecting terms) (a)

II a^2+b^2 = 3b^2-2a^2+1  add 3b^2-2a^2 to both sides 
III 2a^2+2b^2 = 6b^2-4a^2+2  doubling 
IV a^2+b^2 = 5b^2-5a^2+2  add -b^2-a^2 to both sides 
So a^2+b^2 = 5n+2  substitution 
When a^2+b^2+1 = 5n+3  (b) 
    
V 3a^2-2b^2 = 1  from (a)
VI 4a^2-b^2=a^2+b^2+1  add a^2 +  b^2 to both sides 
VII 4a^2-b^2 = 5n+3  substitution, see (b) 
VIII 4a^2-b^2 =(2a+b)(2a-b)   
Since VIII factorises, 5n+3 must be compound, for n>0 .

 

Edited on October 27, 2013, 10:21 am

Posted by broll on 2013-10-27 09:57:07