Week 7

So last week (or the week before that) we learned how to multiply one number by another number

But in all seriousness, we learned that if we have a number of the form $2M^2-1$ where $M$ is an integer, then $2\left(M+K\right)^2-1$ describes it's multiples as long as $K \equiv 0$ or $-2M \bmod 2M^2-1$. By adding $K=Q\left(2M^2-1\right)$ or $K=Q\left(2M^2-1\right)-2M$ we are effectively multiplying the number by $\left(2MQ+1\right)^2-2Q^2$ or $\left(2MQ-1\right)^2-2Q^2$ (Sorry if I changed the name of certain variables).

Let us remind ourselves of the reason we are finding the multiples of $2M^2-1$. We were able to find arrow relations between Plato triples using

$$Plato \rightarrow Plato\\\boxed{\frac{m_{from}}{m_{to}}=\frac{2J^2-1}{2L^2-1}\\i_{from}=L-1\\i_{to}=J}$$

We wanted to know if we could change our values of $J$ and $L$ such that $M_{from}$ and $M_{to}$ don't change. For example, when $J=L$, you can add any number to both

$$\frac{1}{1}=\frac{2\left(1\right)^2-1}{2\left(1\right)^2-1}=\frac{2\left(2\right)^2-1}{2\left(2\right)^2-1}\cdots$$

This is the reason why every Plato triple has a right arrow going to the next term in the Stifel sequence. By finding more of these, we might be able to fully describe the arrow relations between two multiples

So we need to multiply two numbers by the same $\left(2MQ+1\right)^2-2Q^2$ or $\left(2MQ-1\right)^2-2Q^2$

If we say $\left(2MQ+1\right)^2-2Q^2=7$, then we are also saying that $Q\left(2QM^2-Q+2M\right)=3$. Since $Q$ and $M$ are integers we see that the only solution is $Q=1, M=1$. (We started with $7$ because multiplying by $1$ doesn't make sense and the numbers between can never divide a number of the form $2M^2-1$)

If we say $\left(2MQ-1\right)^2-2Q^2=7$, then we are also saying that $Q\left(2QM^2-Q-2M\right)=3$. Since $Q$ and $M$ are integers we see that the only solutions are $Q=1, M=2$ and $Q=3, M=1$

If we interpret these solutions, we see that the only numbers we can multiply by $7$ and $1$ and $7$. (One of those solutions was basically a duplicate) We can see that $2\left(1\right)^2-1=1\rightarrow 7=2\left(2\right)^2-1$ and that $2\left(2\right)^2-1=7\rightarrow 49=2\left(5\right)^2-1$

I will continue exploring this with numbers higher than $7$, but here is a cool pattern I found related to $2M^2-1$. Firstly, here is a list of $2M^2-1$ up to $M=30$

1

7

17

31

49

71

97

127

161

199

241

287

337

391

449

511

577

647

721

799

881

967

1057

1151

1249

1351

1457

1567

1681

1799

We start at $1$ and move forward $\sqrt{1}$ to $7$. $7^2=49$, the next perfect square after $1$.

If we take $49$ and move forward $\sqrt{49}$ to $287$, we might notice that $287=7*41$. And it just so happens that $41^2=1681$ is the next perfect square.

I haven't looked at why this is, but I suspect it is a side effect of $2M^2-1=N^2$ being a Pellian equation.

Another interesting pattern occurs when we consider number we can multiply $2M^2-1$ by

If we consider $M=2$, the first 10 numbers we can multiply by are

7.0

23.0

41.0

73.0

103.0

151.0

193.0

257.0

311.0

391.0

If we consider $M=3$, the first 10 numbers we can multiply by are

23.0

47.0

113.0

161.0

271.0

343.0

497.0

593.0

791.0

911.0

You may have noticed that 23 is second in $M=2$ and first in $M=3$. This pattern actually continues, so hopefully I will be able to exploit it and discover more patterns

This week's calculator will be up shortly

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