Suppose $n^2 + 1$ is a prime number. Then $n + 1$ is also a prime number.

Question

Since this proposition is false, I tried to do a counter-example. Is this a correct format of a counterexample for such proposition?

Take $n = 3$,

$3^2 + 1 = 10$, $10$ is not a prime number as it has factors of $1, 2, 5, 10$.

$3 + 1 = 4$, $4$ is not also prime number as it has factors of $1, 2, 4$.

Thus, this proposition is false.

No, you want to find an integer $n$ such that $n^2 +1$ IS a prime number, but $n+1$ isn't. How do you know that the claim is wrong if I may ask? — Sebastian Schulz, Mar 28 '17 at 04:19
i just tried substituting a number in $n$ for both of the propositions. — User426190, Mar 28 '17 at 04:21

score 3 · Accepted Answer · answered Mar 28 '17 at 04:22

3

In your reasoning for $n=3$ you have that $n^2+1$ isn't prime. Try to fix a $n$ such that $n^2+1$ is prime but $n+1$ is not.

For example:

$n=14 \Rightarrow 14^2 + 1 = 197$ is prime but $14+1=15$ isn't prime since $15=3\cdot5$

answered Mar 28 '17 at 04:22

kub0x

I'm new to propositions and proof writing, and it seems I misread the question. Thought I had to find a $n$ that didn't work for both. – User426190 Mar 28 '17 at 04:26

score 0 · Answer 2 · answered Apr 30 '17 at 09:16

0

The statement is false. For a counter-example, take e.g. n = 14, when n^2 + 1 = 197 which is a prime, and n + 1 = 15 which is not.

answered Apr 30 '17 at 09:16

Han Fred Gu

1

What does this add to kubox's answer? – Jean-Claude Arbaut Apr 30 '17 at 09:39

2 Answers2