1. Geometric Progression

Hello,
This time I have come up with a geometric progression problem.
Three natural numbers are known to form a geometric progression. The second member is known to be greater than the first one with +1.
It is required to find the third member of the geometric progression and this is all that is known about it. So, so far I thought up this: let a, b and c be the members. a = b - 1 as said and from the fact that we are dealing with a geometric progression we get:
$
b^2=(b-1)c
$

which is an equation.. with two unknown variables.
To solve it successfully we need a system of equations, i.e. a second equation, which I do not know where to get from, which is my problem.
The answer is c = 4.

I suspect there might be two solutions for b and a, although they are not our concern.

Thank you.

2. Hello

What's important here is that we are trying to solve $b^2=(b-1)c$ for $b,c$ two natural integers.

If we assume that $b$ is odd then $b-1$ is even and $b^2$ is odd so $\underbrace{b^2}_{\text{odd}}=\underbrace{(b-1)c}_{\text{even}}$ can't hold : $b$ can't be odd.

Now let's assume that $b$ is even. As 2 divides $b$, 4 divides $b^2$. As $b-1$ is odd, 4 doesn't divide $b-1$ hence from $b^2=(b-1)c$ we get that 4 must divide $c$ : $c=4p$ for some $p\in\mathbb{N}$. We can now write $b^2=(b-1)c$ as $b^2=4(b-1)p\Longleftrightarrow b^2-4pb+4p=0$ which is an equation in terms of $b$.

Using the discriminant we get that the solutions of this equation are $b_1=2p+2\sqrt{p(p-1)}$ and $b_2=2p-2\sqrt{p(p-1)}$. We know that $b$ is an integer so we are looking for the values of $p$ such that at least one of the two solutions is an integer. When does this happen ?

3. This happens when p(p-1) is zero, i.e. p = 0 or p = 1. 0 is not a natural number so p = 1 is a solution and c is, indeed, 4.
However, I am unsure how to prove that $p(p-1)=k^2$ can never occur for any integer k bigger or equal to 0.
I have come up with this:
Let $p(p-1)=k^2$, then $p^2-p-k^2=0$. The discriminant is $D=1+4k^2$... which should not be an exact square? It can never equal 4, 9, 16, etc., I mean. Is this so?

I was not expecting this to turn out to be a diophantine equation.

4. Originally Posted by Logic
This happens when p(p-1) is zero, i.e. p = 0 or p = 1. 0 is not a natural number so p = 1 is a solution and c is, indeed, 4.
However, I am unsure how to prove that $p(p-1)=k^2$ can never occur for any integer k bigger or equal to 0.
I have come up with this:
Let $p(p-1)=k^2$, then $p^2-p-k^2=0$. The discriminant is $D=1+4k^2$... which should not be an exact square? It can never equal 4, 9, 16, etc., I mean. Is this so?
Yes, that's it. When I first solved this part of the question I didn't use the discriminant but it works well : The positive root of $p^2-p-k^2$ is $\frac{1+\sqrt{1+4k^2}}{2}$ which is a natural integer iff $1+4k^2$ is a perfect square. Assume $1+4k^2=q^2$ for some $q\in\mathbb{N}$. Rearranging and factoring we get $1=(q-2k)(q+2k)$. As $q-2k$ and $q+2k$ are integers they must both equal 1 hence the only solution is $q=1$ and $k=0$.