I couldn't find an answer to this but I did find a proof that this question is incorrect:
We have $3n – 4$ = prime, so, n cannot be of the form $2q$, $4q$, $8q$, because then I'd be able to factorise it
Similarly, $4n – 5$ = prime, so n is not of the form $5q$
And, $5n – 3$ = prime, so n is not of the form $3q$, $6q$, $9q$
Which means, $n$ is certainly an odd prime.
If we look at $5n – 3$, $5$ and $n$ are both odd, so their product is odd
$3$, when subtracted from an odd number, would always give an even number, which creates a contradiction.
So who's wrong, me or the problem? If I'm wrong, do tell me why and also the solution to the problem, thanks
Best Answer
Notice that the sum of the three primes is $$(3n-4)+(4n-5)+(5n-3)=12(n-1)$$ which is clearly even. Therefore, not all of the primes are odd. It follows that one of them has to be equal to the only even prime, $2$.
Since $n\in\mathbb{N}$, either $3n-4=2$ or $5n-3=2$, implying that either $n=2$ or $n=1$.
$n=1$ is rejected since in this case, $3n-4<0$. $n=2$ gives $2, 3, 7$ as our primes. Therefore, $n=2$ is the only solution to the problem.