SOLUTION: given d|n+1 and d|n^2+1 and d|2 prove that if n is odd then gcd(n+1,n^2+1) = 1

Algebra ->  Divisibility and Prime Numbers -> SOLUTION: given d|n+1 and d|n^2+1 and d|2 prove that if n is odd then gcd(n+1,n^2+1) = 1      Log On


   

Question 1136860: given d|n+1 and d|n^2+1 and d|2
prove that if n is odd then gcd(n+1,n^2+1) = 1
Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.

Your post seems strange to me, and I'd like to discuss it with you.



This fragment  " d |2 ", in standard context (in standard designations), means that " d divides 2 ",  

or (which is the same),  " d is divisor of 2 ".



Then, since we consider only natural numbers in this assignment, it implies  d = 2.



Then the first line  (the "given" part) means that  n+1  and (n^2+1) both are even numbers.



But it contradicts to what has to be proved.

So, in my view, the post is DEFECTIVE and must be edited.