# We can't be prime together

$\nexists~a,b\in\Bbb Z^+~(a\neq b)~\mid~a+x\in\Bbb P\iff b+x\in\Bbb P~\forall~x\in\Bbb Z_{\geq 0}$

Prove the statement above.

Details and Assumptions :

• $$\Bbb P$$ denotes the set of all prime numbers.
• $$\Bbb Z^+$$ denotes the set of all positive integers.
• $$\Bbb Z_{\geq 0}=\Bbb Z^+\cup \{0\}$$ denotes the set of all non-negative integers.

###### Source : Math StackExchange

Note by Prasun Biswas
2 years, 7 months ago

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Seems simple if you use contradiction.

Suppose there does exist such an $$a$$ and $$b$$. WLOG, let $$a > b$$, and $$a - b = d$$. Now let $$p$$ be the smallest prime such that $$p > a$$.

Then, by induction,we can prove all $$p + nd$$ for all natural numbers $$n$$.

This implies that in every set of $$d$$ consecutive integers, starting from an integer larger than $$p$$, we can find at least one prime.

But this is not true for the set $$k(d+1)! + 2, k(d+1)! + 3, ... k(d+1)! + (d+1)$$, in which all elements are composite, and the first element is larger than $$p$$ by taking some suitable k. Hence, we have a contradiction.

- 2 years, 7 months ago

Yes, that is the intended proof. +1

However, you can cut the proof short by taking $$n=p$$ which would result in us getting that $$p+pd=p(1+d)$$ is prime. Contradiction.

The result follows.

- 2 years, 6 months ago

Hey seniors and respected professionals, please do comment in this note. I see that nowadays the scope of Brilliant has been limited to just JEE syllabus and Olympiad Mathematics, please try to promote Higher Mathematics too.

- 2 years, 7 months ago

This problem might well fit into the Olympiad math style actually. There's a really elegant proof (by contradiction) of this that uses just elementary mathematical induction.

- 2 years, 7 months ago

Really? Cool then..

- 2 years, 7 months ago