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2019-11-13, 05:49
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#12 |
"Matthew Anderson"
Dec 2010
Oregon, USA
14408 Posts |
LaurV,
do not believe that there is "a square-free counterexample" |
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2019-11-13, 20:13
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#13 |
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"Matthew Anderson"
Dec 2010
Oregon, USA
80010 Posts |
This is a joke -
I give you Anderson conjecture # 23. for every square free positive integer r, there does not exist any non-zero integer s and t such that sqrt(r) = s + sqrt(t). I believe I saw a theorem about this. |
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2019-11-14, 10:56
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#14 | |
Dec 2012
The Netherlands
2×23×37 Posts |
Quote:
The key - as LaurV suggested - is that the square root of any integer is either an integer or it is irrational. Take any fraction \(\frac{a}{b}\) where a & b are integers (with b≠0). We can choose them so that gcd(a,b)=1. Suppose \(\left(\frac{a}{b}\right)^2=n\) for some integer n. Then \(a^2=nb^2\) so every prime number dividing b also divides \(a^2\) so it also divides a. But gcd(a,b)=1 so there are no such prime numbers. Hence b is 1 or -1 and the fraction \(\frac{a}{b}\) was an integer all along. |
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2019-11-14, 13:09
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#15 | |
Feb 2017
Nowhere
4,643 Posts |
Quote:
1 = -1 + sqrt(4) If we assume r is a positive integer, and not the square of an integer, and t is a positive integer, we can argue as follows: sqrt(r) = s + sqrt(t) sqrt(r) - sqrt(t) = s r - 2*sqrt(r*t) + t = s2 -2*sqrt(r*t) = s2 - r - t The right side is an integer, hence r*t is the square of an integer. Thus, t = r*v2, v could be rational, but t is an integer [The denominator of v2 must divide the largest square factor of r. If you assume r is square free and r > 1, then v must be an integer]. Then -2*r*v = s2 - r - r*v2 r*v2 - 2*r*v + r = s2 r*(v - 1)2 = s2 The left side, though an integer, is not the square of an integer, because r is not the square of an integer. The right side is the square of an integer. Uh-oh, trouble... |
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