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This is a cute little easter egg. However, it's actually not too hard to disprove 3987^12 + 4365^12 = 4472^12 with just a calculator. The thing to do is concer
by ggchappell 12y ago
This is a cute little easter egg.
However, it's actually not too hard to disprove 3987^12 + 4365^12 = 4472^12 with just a calculator. The thing to do is concern yourself with the rightmost digits, not the leftmost.
The fake solution is actually a bit better than this article indicates, since checking only the rightmost digit -- working modulo 10 -- doesn't show the problem. 7^12 ends in 1, and 5^12 ends in 5. So the value of the left-hand side of the proposed equation ends in 6. And so does 2^12.
But checking the rightmost 2 digits -- working modulo 100 -- we can see that it doesn't work. 87^12 ends in 81, and 65^12 ends in 25. Their sum ends in 06. But 72^12 ends in 16. So the proposed equation is false.
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EDIT. For anyone who wonders how to do this.
Raising a number to the 3rd power, and then raising that to the 4th power, we get the 12th power of the original number. If we are working modulo 100, then we can throw out all but the rightmost 2 digits at any stage.
So, what are the rightmost 2 digits of 3987^12? Start by finding 87^3: 658503. We only care about the "03" at the end. 03^4 = 81. So 3987^12 ends with 81.
Similarly, for 4365^12: 65^3 = 274625. 25^4 = 390625. So 4365^12 ends with 25.
And, for 4472^12: 72^3 = 373248. 48^4 = 5308416. So 4472^12 ends with 16.
81 + 25 = 106, which does not end with 16.
- Chinjut 12y agoAnother simple disproof: 3987 and 4365 are both divisible by 3 (and thus so is the left-hand side), but 4472 is not (and thus neither is the right-hand side, as 3 is prime).
- ggchappell 12y agoVery nice! Another easy one is looking at it modulo 4. The powers on the left-hand side are both odd perfect squares, which are thus 1 modulo 4. So their sum is 2 modulo 4. But the right-hand side is an even perfect square, which is 0 modulo 4. EDIT. I guess my modulo-100 discussion is really just a convoluted way of coming up with the above fact.
- Chinjut 12y agoSo it turns out not only could we ignore all but the last digits of the bases... we could furthermore ignore all but the last digits of the exponents!
- yen223 12y agoEasiest disproof: open a Python console and type in 3987**12 + 4365**12 - 4472**12 Arbitrary-precision integers FTW!
- kps 12y agoHomer³ — the 3D sequence in the Halloween episode Treehouse of Horror VI, three years previous — had a similar ‘counterexample’: 1782¹²+1841¹²=1922¹². And also, 1+1=2, e^iπ=-1, and P=NP. I'm so old I remember when The Simpsons was good.
- _kst_ 12y agoAnd that one is easily disprovable by noting that 1782 is even, 1841 is odd, so any sum of their powers must be odd -- but 1922 is even.