Thread: natural numbers

Hi;

Find all natural numbers for which is a prime number, where stands for the greatest integer that is less or equal to .

Any hint as to where to start appreciated.

If you set where r is 0, 1, 2, 3, or 4.

Here are a couple hints:

Some values of r work and some don't.

If ab=5p, (a>0, b>0, p prime) then either:
a=5 and b=p
a=p and b=5
a=5p and b=1, or
a=1 and b=5p

Post again if you still have trouble. It's a really interesting problem.

Do you mean the floor function ? Like this :

Originally Posted by Bacterius

Do you mean the floor function ? Like this :

Yes, the original poster defined the notation.

I'd rather use the real notation instead of defining a new notation for something that already exists. But yeah, he defined it, but I just wanted to make sure. To use the floor function : \left \lfloor [contents] \right \rfloor.

I guess both notations are considered valid, and yes, I meant the floor function.

Originally Posted by hollywood

If you set where r is 0, 1, 2, 3, or 4.

Here are a couple hints:

Some values of r work and some don't.

If ab=5p, (a>0, b>0, p prime) then either:
a=5 and b=p
a=p and b=5
a=5p and b=1, or
a=1 and b=5p

Post again if you still have trouble. It's a really interesting problem.

Well thanks, not sure if I fully understand what you wrote, I think it may be better to assume that where is since the floor function will further let us get rid of the "remainder" part ( ).

So you would get , where r = 0, 1, 2, -1, or -2. Then .

So this expression must be prime, which means that either p=1 or 5p+2r=1. In the second case, 5p=1-2r and only r=-2 works, but then p=1 and p(5p+2r)=1, which is not a prime.

For the p=1 case, only three of the possibilities for r work, giving you three possible solutions.