## What is the abc conjecture? |

**What is the abc conjecture** | Open and solved problems | Consequences | The algorithm

The greatest common divisor of *a* and *b* must be equal to 1. The number *c* must be equal to *a* + *b*. Take for example *a* = 1. *b* = 8 and *c* = 9. To find out whether this is an abc triple we need to find the integer factorization of these numbers. In this case we have *a* = 1, *b* = 2 × 2 × 2 and *c* = 3 × 3.

Now we'll look for the distinct primes in *a*, *b* and *c* and we multiply them. This is called the *radical* of the triple *a*, *b* and *c*. We can write this as *r*(*a*, *b*, *c*). In our example we have the prime factors 2 and 3. So the radical of the triple 1, 8, 9 is 2 × 3 = 6.

What is the radical of the triple 4, 23, 27?

Your answer:

**Your answer is correct!****Your answer is incorrect.****The correct answer is: 138**

If the radical of a triple is less than *c*, we have an abc triple. We see that 1, 8, 9 is an abc triple as 6 is less than 9.

Another example of an abc triple is *a* = 5, *b* = 27 and *c* = 32. To check if this is a triple, we decompose the numbers into their prime factors. 5 is prime, 27 = 3 × 3 × 3 and 32 = 2 × 2 × 2 × 2 × 2. the radical is r(5, 27, 32) = 5 × 3 × 2 = 30. As this is less than 32. 5, 27 and 32 is an abc triple.

Usually you are less lucky. If you take *a* = 4, *b* = 15 and *c* = 19, the prime decomposition is *a* = 2 × 2, *b* = 3 × 5 and *c* is prime so it's prime decomposition is 19. The radical is 2 × 3 × 5 × 19 = 570, and this is larger than 19.

Which of the following is an abc triple?

*a*. (9, 16, 25)

*b*. (1, 63, 64)

Your answer:

**Your answer is correct!****Your answer is incorrect.****The correct answer is: b**

If you take a random triple of numbers, it's usually not an abc triple, because abc triples are quite rare. There are for example only fifteen abc triples with *c* less than 300. However, there are infinitely many abc triples as *c* gets bigger and bigger. With a little math, we can find infinitely many abc triples.

Have a look at the triples *a* = 1, *b* = 9^{n}-1, *c* = 9^{n}. If we let *n* be 1, 2, 3, ..., we'll get infinitely many triples. Now we'll need to prove those are all abc triples. It is clear that *a* + *b* = *c* and *a* and *b* have no common divisor greater than one. It is also true that *b* is divisible by 8, but this is a little more complicated to see. For *n* = 1 it's true: *b* = 8 and that is divisible by 8. We can write *c* as 8 + 1. If we multiply it by 9 we get the *c* from the next triple (where *n* = 2) and we can write it as 9 × 8 + 9 = 10 × 8 + 1. Again *c* is a multiple of 8 plus 1 and *b* is divisible by 8. We can continue doing this. If *b* is equal to 8 × *m*, *c* equals 8 × *m* + 1. The next *c* is 9 × (8 × *m* + 1) = 9 × 8 × *m* + 9 = 9*m* × 8 + 8 + 1 = (9*m* + 1) × 8 + 1. So the next *b* is (9*m* + 1) × 8, which is divisible by 8.

Now we'll write *b* as 2^{3} × *m*. The distinct prime factors in *b*, are all the distinct prime factors in *m*, and, if 2 isn't one of them, also 2. Some primes can occur multiple times in *b*, but the product of prime factors isn't greater than 2 × *m*. There are no prime factors in *a* and *c* has only one prime factor, 3. So the radical of the triple is not greater than 2 × *m* × 3 = 6 × *m* and this is less than *c* = 8 × *m* + 1. So for every value of *n*, the triple *a* = 1, *b* = 9^{n} - 1, *c* = 9^{n} an abc triple.

We would like to compare triples and call one triple better than another one. We can do this by defining a quality for abc triples in the form of a number. The higher the quality triple, the better the abc triple is. The quality of a triple is called *q* and depends on the number *c* and the radical *r* of the triple: *q* is the power you'll need to raise *r* to to get *c*, so *r*^{q} = *c*. We can also express this with logaritms: *q* = log(*c*)/log(*r*). The smaller the radical compared to *c*, the greater the quality of the triple.

When you have an abc triple, the radical is always less than *c*, so *q* is always greater than 1. Most of the time, the quality is not much greater than one. With our first example *a* = 1, *b* = 8 en *c* = 9 we found r = 6, so *q* = log(9)/log(6) = 1.22629.

We already know there are infinitely many abc triples, all with *q* > 1. But we do not know if there are infinitely many triples with *q* > 1.5. Or, if this isn't the case, maybe there are infinitely many with *q* > 1.1? Or perhaps *q* > 1.01 or 1.001? We don't know this yet. No method is known that can find infinitely many triples for such a *q*.

Now that we know all about abc triples, we can have a look at the abc conjecture. There are two versions of the conjecture, a weak one and a strong one. We don't know if either of them is true, but if the strong one is true, so is the weak one. The other way around is not the case -- That's why we call them weak and strong.

After years of searching for abc triples, the best triple found so far has *a* quality of approximately 1.63. No triple with a higher quality is known. It's not hard to imagine that there is an upper bound for the quality of abc triples and this is exactly what the weak abc conjecture says. According to this conjecture there is a number *g* so no abc triple has a quality greater than *g*.

The strong version of the abc conjecture is about the number of abc triples with a high quality: this number should be finite. Even if you take a number *h* that's only a little bit greater than 1, say *h* = 1.0001, then there are still finitely many triples with a quality greater than *h*, according to the strong conjecture. In other words: for any *h* greater than 1, all of the infinitely many triples have a quality between 1 and *h*, except for a finite number of exceptions.

[Warning: this only works for numbers of at most 52 bits or 15 digits.]

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