Math Help - natural numbers

1. natural numbers

Hi;

Find all natural numbers $n$ for which $\left[\frac{n^2}{5}\right]$ is a prime number, where $[x]$ stands for the greatest integer that is less or equal to $x$.

Any hint as to where to start appreciated.

2. If you set $\left[\frac{n^2}{5}\right]=p\text{, then }n^2=5p+r$ 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.

3. Do you mean the floor function ? Like this :

$\left \lfloor \frac{n^2}{5} \right \rfloor = p$

4. Originally Posted by Bacterius
Do you mean the floor function ? Like this :

$\left \lfloor \frac{n^2}{5} \right \rfloor = p$
Yes, the original poster defined the $\left[x\right]$ notation.

5. 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.

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

Originally Posted by hollywood
If you set $\left[\frac{n^2}{5}\right]=p\text{, then }n^2=5p+r$ 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 $n=5p+r$ where $r$ is $0, 1, 2, -1, -2$ since the floor function will further let us get rid of the "remainder" part ( $\frac{4}{5}<1$).

7. So you would get $n^2=25p^2+10pr+r^2$, where r = 0, 1, 2, -1, or -2. Then $\left\lfloor\frac{n^2}{5}\right\rfloor=5p^2+2pr=p( 5p+2r)$.

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.