I would like to start with a shoutout to Mathologer on YouTube for the inspiration for this first post, of which I have chosen a doozy. As well, I would like to apologize for the web design, I will be working on it over time.

The math question I would like to explore is that of the Sum of Cubes video:

He shows that if you take certain numbers within the powers of one and two and arrange them in a certain order, then the numbers equal each other.

For the first power, the integers, those numbers start with:

1+2=3

4+5+6=7+8

9+10+11+12=13+14+15

And so on. The squares do the same thing:

3^{2}+4^{2}=5^{2}

10^{2}+11^{2}+12^{2}=13^{2}+14^{2}

However, as he points out, the pattern breaks down if you try and extend this into higher dimensions. However, an exploration of these number patterns shows that you can extend into the fourth dimension, if you adjust expectations slightly.

To start, we break down the two existing patterns. As Burkhard mentions in his video, the integers are formed by taking the sum of numbers up to an expected value (n) and multiplying by two. This sum we will call alpha, or a.

Meaning we can construct any line by plugging in values of a. For example if a = 10, then n = 4 and the line becomes:

16+17+18+19+20=21+22+23+24

For the squares, you also end up perfectly when you select a number 4a

36^{2}+37^{2}+38^{2}+39^{2}+40^{2}=41^{2}+42^{2}+43^{2}+44^{2}

But notice that the pattern is already breaking at this point, as the integers contain every single possible positive whole value, where the squares skip a few each time.

So, what happens when you try to extend to the third dimension? The logical extension for the math is 6a, meaning the first and most simple equation should be:

5^{3}+6^{3}=7^{3}?

But that doesn’t work. The equation itself has become off. But that doesn’t mean there isn’t a pattern in play. Specifically, the equation is off by 2. or to be more exact, 2a^{2}.

a = 1, 6a = 6 => 5^{3}+(6a)^{3}=7^{3}+2a^{2}

a = 3, 6a = 18 => 16^{3}+17^{3}+(6a)^{3}=19^{3}+20^{3}+2a^{2}

In each case, the remainder of the equation is lower than starting value.

For the fourth dimension, the remainders are greater than the starting value:

a = 1, 8a = 8 => 7^{4}+(8a)^{4}=9^{4}+64a^{3}

a = 3, 8a = 24 => 22^{4}+23^{4}+(8a)^{4}=25^{4}+26^{4}+64a^{3}

The fifth dimension, as the name suggests, is going to take things into strange territory. The pattern no longer seems to be as simple as a relationship to alpha. So, we first find what the remainders of the equations are:

a = 1, 10a = 10 => 9^{5}+10^{5}=11^{5}+x => x = 2002

a = 3, 10a = 30 => 28^{5}+29^{5}+30^{5}=31^{5}+32^{5}+x => 162048

These numbers make little sense in the scheme of things, because we should expect them to be integers to the fourth Power. However 2002 is factored as 2*7*11*13, making it exceedingly primitive. 162048 itself is 2^{8}*3*211, and it continues with no clear pattern emerging. At least, not one that our three dimensional monkey brains can easily understand.

But what if we could try and glean a pattern from the madness? We construct a 5th dimensional graph using replications of lower dimensional graphs or constructs, work out the locations of the added parts based on coordinates, and try and find the holes? Then we could see the potential shape or shapes of the gaps and see if they make more sense.

That question, however will have to wait until I can truly devote some time to this, and I have taken enough as it is.

Shout out to the fact that others found the same patterns and were stumped at around the same place. (Read through Mathologer’s notes)