Trio Of Primes

Notice that $(3n-4)+(4n-5)+(5n-3)=12n-12$ which is always even for all natural numbers $n$ . Since $4n-5$ is always odd for all natural numbers $n$ , then either $3n-4$ or $5n-3$ must be even. But the only even prime (natural) number is 2.

So, if $3n-4=2$ , $n=2$ . Thus $4n-5=3$ and $5n-3=7$ , which are all prime. Thus, $n=2$ is a solution.

If $5n-3=2$ , $n=1$ . Thus, $4n-5=-1$ and $3n-4=-1$ , which do not contain a factor other than $\pm 1$ . So, $n=1$ is not a solution.

Therefore, there is only $\boxed{1}$ natural number $n$ which satisfies the problem.

For (3n-4), (4n-5), (5n-3) to be prime, all of them must be odd except for prime number 2. This is only in the case of n=2.

Hence, there is only $\boxed{1}$ natural number which satisfies the condition.

Except 2 which follows by mere obsevation.......

It can't be even.... 2 will be factor of first....

It can't be odd.....again 2 will be factor of third.......

I used a parity based approach, like many others.

First, assume $n$ is odd. If $n$ is odd, then $5n-3$ will be an even number. The only even prime is 2, so we can solve for the corresponding $n$ value.

$5n-3=2 \\ 5n = 5 \\ n = 1$

According to the third expression, the only odd value for $n$ that could work is 1. This yields -1 for the two other expressions, which is not prime. Thus, there are no odd solutions for $n$ .

Next, assume $n$ is even. If $n$ is even, $3n-4$ will be even. As before, we can set this expression equal to 2, the only even prime number, and solve for $n$ .

$3n-4=2 \\ 3n=6 \\ n=2$

According to the first expression, the only even value of $n$ that could work is 2. We can test this value in the other two expressions to determine if this indeed works.

$4 \cdot 2-5=3 \\ 5\cdot 2-3=7$

Both of these values are prime, so $n=2$ is a solution.

So, we have shown that:

There are no odd solutions. $n=1$ would be the only possible odd solution (by parity of expression 3), but it does not work with the other expressions.

There is only one even solution. $n=2$ is the only possible even solution (by parity of expression 1), which does work with the other expressions.

Thus, there is only one natural number that fits the description.

n=2: 3(2)-4 = 2 4*(2)-5 = 3 5(2) -3 = 7. 2,3, and 7 are all prime, so 2 works.

3n-4 is even for all even n. 5n-3 is even for all odd n.

Therefore, 2 is the only possibility for n.

If n is odd, 3n-4 and 4n-5 would be odd while 5n-3 would be even. the only even prime is 2, so 5n-3=2 >n=1. if n=1, 3n-4 will be negative so n is not 1. if n is even, 3n-4 would be even while both 4n-5 and 5n-3 would be odd. Again, the only prime even number is 2, so 3n-4=2>3n=6>n=2. when n=2, 3n-4=2, 4n-5=3,5n-3=7 which are all prime. Thus, the number of solutions=1.

Simply, every triplet has one even number, since only one prime triplet exists with an even number $(2,3,7)$ for $n=2$ , the answer is $1.$

Consider two cases: Even number And Odd number Now in even number,consider two more cases 2 and other even number. For 2 the condition is satisfied. For any other even first term is any prime not equal to 2 and hence composite For odd,consider two cases,i.e. 1 and any other odd number. For 1 the first two terms are negative. For any other odd,last term is even not equal to two. Hence,there exists only a solution,i.e. """2""""

For every odd prime above 2, the 3rd term is even( ODD - ODD = EVEN). Thus all primes above 2 do not provide any solution. Now we only need to check for n=2, which gives us the answer.

I just started counting from 1 to 10 and then up to 20 and I only found 1, which is 2