# For any $a \in \Bbb{Z}$, can we always find two prime numbers $p, q$, such that $p - q \in (a)$? [duplicate]

This is a major weakening of many prime sum / difference existence questions.

Let $$a \in \Bbb{Z}$$ and $$(a)$$ the ideal generated by $$a$$. Then do there exist two primes $$p, q$$ such that $$p - q \in (a)$$ at least?

Thanks.

## marked as duplicate by Bill Dubuque abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 2 '18 at 19:27

• Sure. Pick a prime $q$ not dividing $a$. Then Dirichlet tells us there are infinitely many primes $p\equiv q \pmod a$. – lulu Nov 1 '18 at 15:31
To see this, note that there are infinitely many primes, but only finitely many remainders on division by $$a$$. By the pigeonhole principle, there are two primes $$p$$ and $$q$$ with the same remainder, and we are done.