Episode 11: Series Finale

Last week I gave you these three equations:

1. 2x²-5y²=5z²-1

2. (x²-21)(x²+2)=y⁹

3. a²+b²=11c²

I will solve them now.

1. 2x²-5y²=5z²-1

First, let's rewrite the equation.

2x²-5y²-5z²=-1

Now, let's look at the equation modulo 5.

2x²=-1 (5)

Now, there is this theorem, called Euler Fermat, that says a^(p-1)/n=0,1, or -1 modulo p. Therefore, since 2=(5-1)/2, we see we have three cases.

2(0)=-1 (5) (i)
2(1)=-1 (5) (ii)
2(-1)=-1 (5) (iii)

Now, if we simplify we can see,

0=-1 (5) (i)
2=-1 (5) (ii)
-2=-1 (5) (iii)

Since none of these are true, we can say that there are no solutions to this equation. ♥

2. (x²-21)(x²+2)=y⁹

There is a bit of complicated math in this solution, so I'm just going to show you the end of the solution. Let's look at:

x²-21=y₁⁹
-x²+2=y₂⁹

So we have

23=y₂⁹-y₁⁹

So 23 is the difference of two powers of 9. But let's list the powers of 9:

0, 1, huge

There is no way that 23 is the difference of two powers of 9. Therefore, there are no solutions to this equations. ♥*

3. a²+b²=11c²

There is this great theorem that says that the sum of two square numbers cannot divide a prime of the form 4k-1. We can see that 11 is a prime of the form 4k-1 (where k=3), and (a²+b²) is the sum of two squares. Therefore the left-side cannot divide the right-side and thus there are no nontrivial solutions. (I say nontrivial because a=0, b=0, and c=0 obviously works). ♥

Now, to end this wonderful series, I will share with you one of my favorite theorems and its proof from this semester. Please don't be concerned if you don't understand.

Theorem: Prove that a Hilbert space contains an orthonormal basis.

Proof: ZORN! ♥

I hope you have enjoyed reading these
as much as I have enjoyed writing them!!

*The solution is a bit more difficult. It requires knowledge of the Power Product Lemma and Quadratic residues. But the proof basically boils down to that.