http://www.sciencenews.org/articles/20060610/bob8.asp Please Register or Log in to view the hidden image! yale
They mention that 1782^12 + 1841^12 = 1922^12 can easily be seen to be false by an even/odd argument, and they later used: 3987^12 + 4365^12 = 4472^12 which isn't susceptible to this attack. I claim this equation can be seen false with no more than a few seconds of mental computations, in at least two (kinda similar) ways. How is this so? (No invoking Fermat's, no mechanical aids)
4365^12 ends with the '5' digit and 3987^12 ends in a '1' and so... hmmm.. nope. i wonder if the # of digits for 4472^12 is not enough. ok. mathematica time. Please Register or Log in to view the hidden image! just kidding.
Mod 3 arithmetic will do it, if I haen't made a mistake: 3987^12 = 0 mod 3 4365^12 = 0 mod 3 4472^12 = 1 mod 3
I could be mistaken, but here goes: The least significant digit in a multiplication is not affected by any digits to the left of it. 9 times 9, keep the LSD, drop everything to the left, equals 1 no matter what is to the left. Thus it becomes obvious that 3987^12 ends in 1. Multiply by pairs. You have six pairs ending in 1. It doesn't matter how many 1s you multiply, that result doesn't change. 4365^anything is going to end in 5. The sum of 3987^12 and 4365^12 ends in 5 plus 1, which is six. 2 to the 12th power is 4096. This isn't disproven by adding the last digits. I suppose that it is a valid counterexample if proven even if the two numbers are also perfect squares, because they are cubes, 4th, 6th, and 12th powers.
Your work is correct, so is your conclusion. Considering the last digits (essentially working mod 10) won't do, but good try. Pete's was a simple way I was thinking of, you can work those out in your head very fast with the divisibility by 3 test, 3987 is divisible by 3 iff the sum of it's digits are (not that dividing by 3 would take long), etc. Left hand side is divisible by 3, right hand side isn't. Mod 4 is also very quick. the square of any odd number mod 4 is 1, the square of any even number gives 0, so we have (odd^6)^2+(odd^6)^2=(even^6)^2 which bceomes 1+1=0 mod 4. Related is the fact that any primitive pythagorean triple x^2+y^2=z^2 must have z even, one of x, y even and one odd (primitive means no common divisors among x, y, z). Lots of other choices of modulus will work, but I don't think any quite so simply.
Divisibility by 3 is conclusive. Two numbers not divisible by three can be added to create a sum that is divisible by three, but two numbers that are can only create sums that are divisible by 3. Amazing what you can do without actually working out all the digits of something like that.
It's the setup for a lot of bad math jokes, but they usually end in a mathematician proving the existance or nonexistance of something, but never actually giving the solution.
Proofs can only be understood if you understand the language. Some are a lot more easily shown to be mathematically precise without working them through than others.
Well usually something is proven the way that is easiest to the person proving it. If it's easier for you to open mathematica and raise expand the equation and check it's truth value, then that is good enough for you. But it's equally valid and easier for me to see do the even/odd argument.
You can approach the mod 3 way the same way as the mod 4 way. If m is not zero mod 3, then m^2=1 mod 3. Considering the optons, if x^2+y^2=z^2 then either 3 divides x, y, and z, or 3 divides either x or y (not both) and does not divide z. This can be extended using some elementary number theory (or elementary group theory if you like). If p is prime and p does not divide m then m^(p-1)=1 mod p. Applying this to x^(p-1)+y^(p-1)=z^(p-1), where p>2, we get a similar conclusion, either p divides x, y, and z, or it divides either x or y (not both) and does not divide z. ^12 we have some options for suitable p's. x^12=(x^6)^2 for p=3 x^12=(x^3)^4 for p=5 x^12=(x^2)^6 for p=7 x^12 for p=13 For 3987^12 + 4365^12 = 4472^12, p=3 is what we did above, though this is a slightly different view. p=5 leads nowhere. p=7 and p=13 do the trick, but divisibility tests for these guys are a little trickier than the p=3 version. Of course this is all unnecessary given Fermat's, but it serves as an example of the kind of things you can do to Diophantine equations with some simple modular arithmetic to learn something about possible solutions.
