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Try #4 on the exercises at the bottom of this website...
http://www.math.rutgers.edu/~erowland/m … metic.html
4. Show that if x, y, z are integers such that x^3 + y^3 = z^3, then at least one of them is divisible by 7.
Yeah..good luck with that!
Last edited by Fruityloop (2009-08-20 11:27:01)
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Hi Fruityloop;
Is it a joke?
I used to know a teacher who claimed he always put mistakes in his problem answers to see if his students were aware. I think he was just covering his rear.
Last edited by bobbym (2009-08-20 20:28:08)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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This is a legitimate exercise. The cubes of 1 to 6 (mod 7) are just 1 and 6.
There's no way to add two of those together to produce 1 or 6 (1+1 = 2, 1+6 = 0, 6+6 = 5), and so if the first equation is true then one of them must be a multiple of 7.
If we're staying in the positive integers then FLT makes it true vacuously, but there's no need to go that deep.
Why did the vector cross the road?
It wanted to be normal.
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Hi;
Is it a joke?
When I first looked at this I just thought about positive integrs so the solution is trivial. Everybody knows that 2 integer cubes cannot sum to a third cube other than for the trivial cases. But when you think about negative integers it is not so well known as to whether the difference of 2 integer cubes can be a cube. So in that sense it is a legitimate question.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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