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Prove that the only prime triple is $3,5,7$.
I tried proving using this method: Multiplication of $3$ jumps back and forth between being an even and an odd number. Thus goes from odd to odd over an interval max size 6, and likewise from even to even. In either case these will have the following types: (even - odd - even - odd - even - odd) and (odd - even - odd - even - odd - even), with 3 being a multiple of the first instance in the combinations. In the both combinations, the first and fourth would be divisible by three, leaving only two possible primes left (two odd numbers). Thus the only triple prime can be a combination where three is considered a prime....
However, is this a formal proof? can anybody tell me how to do a proof for this questions?

user1090614
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    You want to show that one of $x,x+2,x+4$ is always divisible by $3$. That is what your argument is getting at. It could have been done more efficiently: $x$ leaves remainder $0$, $1$, or $2$ on division by $3$. If it is $0$, we are finished with this part of the argument. If it is $1$, then $x+2$ is divisible by $3$. If it is $2$ then $x+4$ is. So one of our numbers must be $3$. – André Nicolas Nov 02 '13 at 21:39
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    Here is a link to a question that asks the statement in André's comment. – Jay Nov 02 '13 at 21:47

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