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

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.

## Comments

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TopNewestSeems 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. – Siddhartha Srivastava · 1 year ago

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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. – Prasun Biswas · 1 year ago

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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. – Swapnil Das · 1 year ago

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– Prasun Biswas · 1 year 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.Log in to reply

– Swapnil Das · 1 year ago

Really? Cool then..Log in to reply