**Fermat's Last Theorem from a ones-digit perspective**

For what it's worth - no stunning insights here, just something I want to get off my chest. There are many pages to find the bio of this remarkable man, Pierre de Fermat, but here we'll concentrate on the world famous theorem.

Fermat's Last Theorem states that

x` ^{n}` + y

has no non-zero integer solutions for x, y and z when n > 2. Fermat wrote

*I have discovered a truly remarkable proof which this margin is too small
to contain. *

...and after reading a couple books on the subject I got to feeling like I knew what the proof was! Or at least that I had a fairly good idea. It occurred to me as I was reading about how Euler actually proved it for n=67, n=73; apparently, focusing on the prime number powers, but unable to tie it all together into a continuous proof. Anyway, this is what I think Fermat saw:

**Part I.** If you multiply 2 by itself over and over, you get the following
sequence:

2,4,8,16,32,64,128,256,512,1024,2048,4096, etc.

Should be familiar to students of binary numbers. If you look at the ones digit you'll notice a repeating pattern: 2,4,8 and 6 keep showing up. If you keep going the multiplication gets steadily more difficult, but the ones digits will always be predictable. For 3, the cycle is 3,9,7,1. Numbers ending in 0,1,5 or 6 stay the same. (see Fig. 1 below)

**Part II.** On one Halloween episode of the Simpsons (Homer^{3})
there was the following equation:

1782^{12} + 1841^{12} = 1922^{12}

Also something you can easily find references to through Yahoo. This equation works on most calculators, which says more about calculators than about the theorem. Using our method, however, rather than grinding through the whole calculation we can quickly see that it doesn't work, because 7 doesn't equal 6. Okay, here goes...

1) Discarding the extraneous digits, we end up with 2^{12} +
1^{12} = 2^{12} .

2) We deal with the powers next. The remainder of 12 / 4 is zero. The formula
doesn't work with powers of zero, so we convert it to 4 instead. From a ones
digit perspective the power of 4 behaves like 12. We now have 2^{4}
+ 1^{4} = 2^{4} .

3) At this point we could subtract 2^{4} from both sides and
get 1^{4} = 0. But if we follow through, 2^{4}
is 16, and 1* ^{n}* = 1, so we have 16+1 = 16, or 17 = 16. If we
keep only the ones digits it becomes 7= 6, which is not true. So, three different
ways of showing it's not true.

1782^{12} + 1841^{12} = 1922^{12}

2^{12} + 1^{12} = 2^{12}

2^{4} + 1^{4} = 2^{4}

16 + 1 = 16

6 + 1 = 6

7 = 6

This method can save us time by showing us if the equation is definitely false
or not. If the ones digits add up, then the equation might be true. For example,
let's deconstruct 3^{5}+10^{2} = 7^{3}.
We change 3^{5} to 3^{1}, and 10^{2}
to 0* ^{2}*, and 7

3^{5}+10^{2} = 7^{3}

3^{1}+0^{2} = 7^{3}

3 + 0 = 343

3 + 0 = 3

3 = 3

**Part III.** Tying all this mess together.

Fig.1 - Ones digit cycles. An integer with a ones digit of 'd' multiplied by itself several times will have a unique ones digit cycle. Numbers ending in 0,1,5, or 6 will stay the same.

Fig. 2 - Ones digit solution grid for powers of 1,5,9,13,17, etc.

Fig. 3- Ones digit solution grid for powers of 2,6,10,14,18, etc.

Fig. 4- Ones digit solution grid for powers of 3,7,11,15,19, etc.

Fig. 5- Ones digit solution grid for powers of 4,8,12,16,20, etc.

The following figures are designed to make this a little easier to understand, if their descriptions don't. This is the groundwork for a proof showing Fermat's theorem is true; I'll leave it to someone else to go the rest of the way. This method breaks the problem down into 400 sub-problems, 96 of which are already done for us. The squares in red in figures 3 and 5 show problems that have no solutions because no integers raised to the respective powers can have those digits for a ones digit. Here's an example.

11^{6} + 11^{6} = ?.

This is not a solution because the answer is 3,543,122 (or 12.347083^{6}),
and there is no integer that can be raised to the 6th power that has a ones
digit of 2. This is also true for powers of 2,10,14, 18, to infinity.

Unfortunately this is not the case for the odd powers. It's all up for grabs, so I need a formula for the other 304 subproblems.

by Constantine Constantinople Jr.