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
xn + yn = zn
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:
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 (Homer3) there was the following equation:
178212 + 184112 = 192212
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 212 + 112 = 212 .
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 24 + 14 = 24 .
3) At this point we could subtract 24 from both sides and get 14 = 0. But if we follow through, 24 is 16, and 1n = 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.
178212 + 184112 = 192212
212 + 112 = 212
24 + 14 = 24
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 35+102 = 73. We change 35 to 31, and 102 to 02, and 73 stays the same. We end up with 3 + 0 = 343. Discarding digits again, we get 3 + 0 = 3, which is 3 = 3. So, we now know the equation MAY BE right. As it is, this one's easy to calculate so we know it's right.
35+102 = 73
31+02 = 73
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.
116 + 116 = ?.
This is not a solution because the answer is 3,543,122 (or 12.3470836), 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.
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