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:
$\displaystyle b^2=(b-1)c$
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 $\displaystyle b^2=(b-1)c$ for $\displaystyle b,c$ two natural integers.

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

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

Using the discriminant we get that the solutions of this equation are $\displaystyle b_1=2p+2\sqrt{p(p-1)}$ and $\displaystyle b_2=2p-2\sqrt{p(p-1)}$. We know that $\displaystyle b$ is an integer so we are looking for the values of $\displaystyle p$ 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 $\displaystyle p(p-1)=k^2$ can never occur for any integer k bigger or equal to 0.
I have come up with this:
Let$\displaystyle p(p-1)=k^2$, then $\displaystyle p^2-p-k^2=0$. The discriminant is $\displaystyle D=1+4k^2$... 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 $\displaystyle p(p-1)=k^2$ can never occur for any integer k bigger or equal to 0.
I have come up with this:
Let$\displaystyle p(p-1)=k^2$, then $\displaystyle p^2-p-k^2=0$. The discriminant is $\displaystyle D=1+4k^2$... 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 $\displaystyle p^2-p-k^2$ is $\displaystyle \frac{1+\sqrt{1+4k^2}}{2}$ which is a natural integer iff $\displaystyle 1+4k^2$ is a perfect square. Assume $\displaystyle 1+4k^2=q^2$ for some $\displaystyle q\in\mathbb{N}$. Rearranging and factoring we get $\displaystyle 1=(q-2k)(q+2k)$. As $\displaystyle q-2k$ and $\displaystyle q+2k$ are integers they must both equal 1 hence the only solution is $\displaystyle q=1$ and $\displaystyle k=0$.