# 1782^12 + 1841^12 = 1922^12?

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Hey everyone... For one of my courses at university (Introduction to Linear Algebra) our teacher gave us a challenge. He said if we can prove the following equation right we get a free book on Linear Algebra, and get to describe how we did it in front of the class. He mentioned that it was a problem that was outstanding for quite some time, and that it was only recently solved. The equation (as far as I can tell what it is): 178212 + 184112 = 192212 Now I am not asking for an answer, but what I wouldn't mind is a point in the right direction. Is there a particular name for this problem? I am guessing wikipedia will no doubt have something on it, but seeing as I don't know what it is called, it is a little hard to find. Any ideas/hints/thoughts?

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I'm not quite sure what you mean by prove it right; they're just numbers, and any calculator will tell you they're the same. What are you meant to be doing?

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Well, considering the size of the numbers, an average calculator wouldn't be able to do it without at least losing some precision.

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Quote:
 Original post by SymphonicIt ain't so

Many thanks and a Rate++ to you, good sir!

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Quote:
 Original post by Moe178212 + 184112 = 192212

False. Not only because of Fermat's Last Theorem, but easily shown for this example:

178212 must have a units digit the same as that of 212 = 4096.
Similarly, 184112 must have a units digit of 1.

So, the units digit of 178212 + 184112 is 7, but that of 192212 is 6.

Thus the equation is false.

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If you only need to prove that one equation then write a computer program to just calculate it. Make a special class that stores numbers as arrays of digits 1 to 10 (or you could save space by using bytes and converting all numbers to base 2). So the number 1922 would be an array of size 4 with the digits 1 9 2 and 2.

Course, for your teacher to be completely satisfied, you'll also have to write up some sort of proof that shows your algorithms and class are correct!

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Quote:
 Original post by rprellerIf you only need to prove that one equation then write a computer program to just calculate it. Make a special class that stores numbers as arrays of digits 1 to 10 (or you could save space by using bytes and converting all numbers to base 2). So the number 1922 would be an array of size 4 with the digits 1 9 2 and 2.Then write overloads for addition, mult, and equality.Course, for your teacher to be completely satisfied, you'll also have to write up some sort of proof that shows your algorithms and class are correct!

Or use Python!

>>> (1782 ** 12) + (1841 ** 12)
2541210258614589176288669958142428526657L

>>> 1922 ** 12
2541210259314801410819278649643651567616L

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Heh, I never knew that Python could do such things! I think I will eventually have to learn Python...

bakery2k1: I am not entirely sure that I follow you. What do you mean by "unit digit"?

(Please, forgive my ignorance of math. I haven't taken a math course sinse high school, and that was over 4 years ago [sad]).

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he's referring to the last digit of the number (one's place, units place- same thing)

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The left side is odd, the right side is even.

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 Original post by alvaroThe left side is odd, the right side is even.

i second that.
[With fermat last theorem, you'll also need to verify a (many pages long) proof to be actually sure.]

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 Original post by Moebakery2k1: I am not entirely sure that I follow you. What do you mean by "unit digit"?

The last digit of the number, i.e. the 1 in 5671, as opposed to the "tens" digit or the "hundreds" digit.

I am basically explaining _why_ alvaro and Dmytry can say that the left side is odd and the right side is even.

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Quote:
Original post by bakery2k1
Quote:
 Original post by Moebakery2k1: I am not entirely sure that I follow you. What do you mean by "unit digit"?

The last digit of the number, i.e. the 1 in 5671, as opposed to the "tens" digit or the "hundreds" digit.

I am basically explaining _why_ alvaro and Dmytry can say that the left side is odd and the right side is even.

Actually, you don't really need to consider digit powers, just using
x^n is divisible by prime p if and only if x is divisible by prime p. I.e. for 2 x^n is odd if x is odd and x^n is even if x is even. You instantly see from equation that it's even+odd=even , it can't be true.

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Would it make any sense if the last power was ^13? I was trying to read it off an image.

In case anyone wants to know where it is coming from, it is from the Simpson's episode where Homer goes into the 3rd dimension (by crawling behind a bookshelf to avoid Patty and Selma). One of the equations in the 3rd dimension is what I previously posted. Our teacher snapped a picture of the equation and posted it on WebCT for all to take a look at.

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Quote:
 Original post by MoeWould it make any sense if the last power was ^13? I was trying to read it off an image.

No. Left side it still odd, and right side is still even.

Besides, look at the values Simian Man posted halfway up. The values of the left and right side are almost the same, just a tiny relative difference (the difference is in the 10:th digit from the left). Changing the power to 13 instead on the left side will make the left side 1922 times larger.

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Just because I can: in Rowan ;D ...

+/¨((1782 1841) 1922)*12
(2.54121025861459E+39 2.5412102593148E+39`)