Fermat's Last Theorem
Andrew Wiles, pictured above, was recently awarded the 2016 Abel Prize for his 1994 proof of Fermat's last theorem .
Suppose n = 2 0 1 6 and ∣ x ∣ , ∣ y ∣ , ∣ z ∣ are all less than 10. How many solutions in integers are there to the equation on the chalkboard?
Note: A solution here refers to an ordered triple ( x , y , z ) .
The answer is 73.
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2 solutions
Damn. Had I seen this picture with the xyz != 0, I probably would have had a better chance at this problem rather than reading it straight out that it says there are no integer solutions.
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@Nick Benallo That makes sense... I mainly just thought it was funny that all of the articles reporting this prize have a picture of Wiles in front of an incomplete version of the theorem! :)
Yeah ! Nice problem ! but this ought to be level 5 isn't it ?
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Hmm, well, the solution (given Fermat's Last Theorem, which is obviously Level 5+) is actually reasonable simple. Hopefully someone will post one soon :)
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As the hint suggests, there is no solution, unless xyz=0.
A trivial solution is x=y=z=0.
Apart from this, we have solutions in the form of 0 < |x| = |z| and y=0 or x=0 and 0<|y|=|z|. Due to symmetry, it is enough to determine the number of solutions in the first case and double it (plus add 1 (the trivial all-zeros case)).
As |x|<10, we can choose from the integers {1,2,3,4,5,6,7,8,9} for |x|=|z| (9 values). As we are using absolute value, both x and z can either be positive or negative, so we have 4 groups of cases: +-, ++, -- and -+ .
4 × 9 = 36
36 × 2 + 1 = 7 3 .
(It's quite a shame, that it is a multiple choice question, as I chose 19 (2×9+1), by not considering the negative numbers first time. If I had the 3 tries as normal, I would have got it right the second time.)
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@Zee Ell – Great solution! I was wrong 'cause I forgot the negative cases.
@Zee Ell – Thank u for your hint.
@Zee Ell – Bah! I thought it was 109, because I screwed up and allowed z=0. If only it were still 2015 :P
Why should this be level 5? It is a simple counting problem - maybe combinatorics level 3?
The theoram written is incomplete. It has trivial solutions although n ≥ 3 . Trivial solutions includes that either x or y = 0. Since ∣ x ∣ , ∣ y ∣ , ∣ z ∣ < 1 0 we have − 1 0 < x , y , z < 1 0 we have total 7 3 tuples . ( x , y , z ) = ( 0 , 0 , 0 ) , ( 0 , 1 , 1 ) . . . ( 0 , 9 , 9 ) , ( 0 , 1 , − 1 ) . . . ( 0 , 9 , − 9 ) , ( 0 , − 1 , 1 ) . . . ( 0 , − 9 , 9 ) , ( 0 , − 1 , − 1 ) . . . ( 0 , − 9 , − 9 ) , ( 1 , 0 , 1 ) . . ( 9 , 0 , 9 ) , ( 1 , 0 , − 1 ) . . . ( 9 , 0 , − 9 ) , ( − 1 , 0 , 1 ) . . . ( − 9 , 0 , 9 ) , ( − 1 , 0 , − 1 ) . . . ( − 9 , 0 , − 9 )
Hint: That's not exactly the full theorem... Here's a fixed version:
Note that x y z = 0 is a necessary condition, but not yet sufficient. What are all of the cases that we should consider?