Difference between revisions of "1959 IMO Problems/Problem 1"
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− | Assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]] where <math>p</math> is a [[ | + | Assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]] where <math>p</math> is a [[ |
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<math>21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}</math> | <math>21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}</math> | ||
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Latest revision as of 20:06, 3 March 2021
Contents
Problem
Prove that the fraction is irreducible for every natural number .
Video Solution
https://youtu.be/zfChnbMGLVQ?t=1266
~ pi_is_3.14
Solution
Denoting the greatest common divisor of as , we use the Euclidean algorithm:
It follows that is irreducible. Q.E.D.
Solution 2
Assume that is a reducible fraction where is a [[
Subtracting the second equation from the first equation we get which is clearly absurd.
Hence is irreducible. Q.E.D.
Solution 3
Assume that is a reducible fraction.
If a certain fraction is reducible, then the fraction is reducible, too. In this case, .
This fraction consists of two consecutives numbers, which never share any factor. So in this case, is irreducible, which is absurd.
Hence is irreducible. Q.E.D.
Solution 4
We notice that:
So it follows that and must be coprime for every natural number for the fraction to be irreducible. Now the problem simplifies to proving irreducible. We re-write this fraction as:
Since the denominator differs from a multiple of the numerator by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that is irreducible.
Q.E.D
Solution 5
By Bezout's Lemma, , so the GCD of the numerator and denominator is and the fraction is irreducible.
Solution 6
To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with + . Simplifying the fraction, we end up with. . Now combining these fractions with addition as shown before in the problem, we end up with . It's important to note that 17 is 1 off from being divisible by 6, and you'll see why later down this explanation. Now we expirement with some numbers. Plugging 1 in gives us . Notice how both numbers are one off from being divisible by 8(25 is next to 24, 17 is next to 16). Trying 2, we end up with . Again, both results are 1 off from being multiples of 15. Trying 3, we end up with . Again, both end up 1 away from being multiples of 22. This is where the realization comes in that two scenarios keep reoccurring:
1. Both Numerator and Denominator keep ending up 1 value away from being multiples of the same number, but
2. One always ends up being a prime number. Knowing that prime numbers only factors are 1 and itself, the fraction ends up being a paradoxical expression where one prime number is always being produced, and even with larger values, like say 15, implementing it in gives us , we keep ending up with results that at the end will only have a common divisor of 1.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1959 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |