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The title says it all ... Obviously, any such triple must be of the form $(4a+1,4a+2,4a+3)$ where $a$ is an integer. Has this problem already been studied before ? The result would follow from Dickson's conjecture on prime patterns, which implies that there are infinitely many integers $b$ such that $4(9b)+1,2(9b)+1$ and $4(3b)+1$ are all prime (take $a=9b$).

A related question : Question on consecutive integers with similar prime factorizations

The title says it all ... Obviously, any such triple must be of the form $(4a+1,4a+2,4a+3)$ where $a$ is an integer. Has this problem already been studied before ? The result would follow from Dickson's conjecture on prime patterns, which implies that there are infinitely many integers $b$ such that $4(9b)+1,2(9b)+1$ and $4(3b)+1$ are all prime (take $a=9b$).

A related question : Question on consecutive integers with similar prime factorizations

The title says it all ... Obviously, any such triple must be of the form $(4a+1,4a+2,4a+3)$ where $a$ is an integer. Has this problem already been studied before ? The result would follow from Dickson's conjecture on prime patterns, which implies that there are infinitely many integers $b$ such that $4(9b)+1,2(9b)+1$ and $4(9b)+3$ are all prime (take $a=9b$).

A related question : Question on consecutive integers with similar prime factorizations

The title says it all ... Obviously, any such triple must be of the form $(4a+1,4a+2,4a+3)$ where $a$ is an integer. Has this problem already been studied before ? The result would follow from Dickson's conjecture on prime patterns, which implies that there are infinitely many integers $b$ such that $4(9b)+1,2(9b)+1$ and $4(3b)+1$ are all prime (take $a=9b$).

A related question : Question on consecutive integers with similar prime factorizations