abc Conjecture - Numberphile

The abc Conjecture may have been proven by a Japanese mathematician - but what is it?
More links & stuff in full description below ↓↓↓

NUMBERPHILE

Videos by Brady Haran

[marlenedietrich]

to those interested: the paper didn't get accepted as a proof

edit: based on the replies, it seems that it may have been revisited. So I can't say what the status on it is

[pieguy]

Pieguy159's conjecture:
Whenever you use a marker, you will always end up writing on yourself.

[MewPurPur]

To those interested: a 600-page redo of the proof is gonna be published in a peer-reviewed journal. But it's still not accepted as a proof and even more controversy among mathematicians is going on now.

[aliebadi]

I have discovered a truly marvelous proof of the abc conjecture which my brain is too small to contain.

[SimplyMyAccount]

At 0:51 when Brady accidentally shows the mannequin and then has to show the whole thing so its not weird that theres a face in the frame

[singingbanana]

Fermat was almost certainly mistaken. It can happen to anyone - for example my mistake in this video!

[Gelfling]

"it's called the radical because it is well wicked" omfg im in tears

[singingbanana]

If abc is true then there is an argument that shows there are no examples of x^n + y^ = z^n, if x and y are whole numbers and don't share any factors. That's nearly Fermat's Last Theorem, but the full version isn't restricted to whole numbers etc.

[ExplosiveBrohoof]

4:07 isn't inaudible. He said that "it's called the radical because it is well wicked."

[dansanger]

I have discovered a truly marvelous proof of the abc conjecture that is too large to fit in a YouTube comment. I'll publish it when I have the time.

[numberphile]

Some of out have quite correctly pointed out (to James' horror) that 2*3*5*13*17=6630, not 13260

The explanation still stands, this error withstanding.

[gdogvibes]

Even something this simple needs a 500 page explanation... Amazing how such simple questions have such complicated answers.

[ThorHC]

Funnily enough, Mochizuki's proof is still being checked to this day.

[woodfur]

I love the way his eyes light up when he's talking about how big this could be if it's proven. Dr. Grime's passion for maths really rubs off in his videos in a way that even other Numberphile speakers just can't match. He makes you want to learn it, and I love that.

[numberphile]

There's a good explanation, but I'll leave it as a mystery for a bit longer!

[vimages]

a + b = c! Yes! At last a Numberphile I might actually understand .... or so I thought!

[ROCCOANDROXY]

The conjecture basically states the following:
Definition:
For any positive integer n the radical of n denoted by rad(n) is the product of distinct prime factors of n.
For example rad(12) = rad(2^2 * 3) = 2 * 3 = 6.
Let a, b, and c be positive integers. If gcf(a, b, c) = 1 such that a + b = c, then it is usually the case that rad(a * b * c) > c so the abc conjecture deals with the exceptions.
Now, to state the conjecture.
For every epsilon > 0 there exists only a finite number of triples (a, b, c) with gcf(a, b, c) = 1 and a + b = c such that c > rad(a * b * c)^(1 + epsilon).

Let a = 12 = 2^2 * 3
b = 11
c = a + b = 23
and rad(a * b * c) = 2 * 3 * 11 * 23 > c which is the usual case.

The power (1 + epsilon) is necessary since there are an infinite triples (a.b.c) with gcf(a, b, c) = 1 and a + b = c such that c > rad(a * b * c).
For example:
a = 1
b = 2^(6 * n) - 1, where n is a positive integer
c = 2^(6 * n)
Show: 9|b
For n = 1 we have b = 63 = 7 * 9 --> 9|b
Assume 9|b for some positive integer n > 1 --> 2^(6 * n) - 1 = 9 * j.
Show 9|(2^(6 * (n + 1)) - 1)
2^(6 * (n + 1)) - 1 = 2^(6 * n + 6) - 1 = 2^(6 * n) * 2^6 - 1 = (9 * j + 1) * 2^6 - 1 = 2^6 * 9 * j + 2^ 6 - 1 = 2^6 * 9 * j + 63 =
9 * (2^6 * j + 7) = 9 * k --> 9|(2^(6 * (n + 1)) - 1)
By mathematical induction 9|b for all positive integers n.

Now rad(b) = rad(3^2 * j) = 3 * j = b/3 and rad(c) = 2 --> rad(a * b * c) = 2 * rad(b) = 2 * b/3 < 2 * c/3 < c.

If anyone is interested I wrote a short program(using free pascal) that generates some of the above triples satisfying the abc conjecture.

program test; {for abc conjecture}
uses mathunit;
var testabc:text;

procedure abc;
var a, b, c, prod, j:longint;
const last1 = 50;
last2 = 5000;
last3 = 30;

begin
assign(testabc, 'abc.txt');
rewrite(testabc);
for a:= 1 to last1 do
begin
for b:= a + 1 to last2 do
begin
c:= a + b;
if gcf(gcf(a, b), c) = 1 then
begin
prod:= rad(a) * rad(b) * rad(c);
for j:= 2 to last3 do {abc conjecture}
begin
if c > power(nroot(prod, j), j + 1) then
writeln(testabc, 'c = ', c, ' > rad(a * b * c))^', j + 1, '/', j, ' = ', power(nroot(prod, j), j + 1):0:9, ' where a = ', a, ' b = ', b);
end;
end;
end;
end;
close(testabc);
end;

begin
abc;
readln;
end.

The functions used above where in my unit mathunit. You can make a unit for them or use them directly in the above program.

function prime(p:integer):boolean;
var n:integer;

begin{prime}
if p = 1 then
prime:= false
else
if p = 2 then
prime:= true
else
begin{else}
prime:= true;
n:=2;
while (n < trunc(sqrt(p)) + 1) do
begin{loop}
if p mod n = 0 then
prime:= false;
n:= n + 1;
end;{loop}
end{else}
end;{prime}

function gcf(m, k:LONGINT):LONGINT;
{lcm calls gcf ---> need longint}
var n, factor, min:INTEGER;

begin{gcf}
if m <= k then
min:= m
else min:= k;

for n:= 1 to min do
begin{loop}
if (k mod n = 0) and (m mod n = 0)
then
factor:= n;
end;{loop}

gcf:= factor;
end;{gcf}

function power(base:real; exponent:integer):real;
var n:integer;
product:real;

begin
product:= 1;
for n:= 1 to exponent do
product:= product * base;

power:= product;

function nroot(x:real; n:integer):real;

const error =
var right, left, midpt, diff:real;
count:integer;

begin{nroot}
IF (ABS(X) >= 1) THEN
BEGIN
left:= 1; right:= ABS(x);
END;
IF (ABS(X) < 1) THEN
BEGIN
LEFT:= ABS(X); RIGHT:= 1;
END;

midpt:= (left + right)/2;
count:= 1;
repeat
if power(midpt, n) > ABS(x) then {X CAN < 0 FOR N ODD}
right:= midpt {ASSUMES X >= 0 FOR N EVEN}
else {ALWAYS USE SAFEGUARD FOR N EVEN}
left:= midpt;
count:= count + 1;
midpt:= (left + right)/2;
diff:= abs(ABS(x) - power(midpt, n));
until((diff < error) or (count = 100));

IF (X < 0) AND (N MOD 2 = 1) THEN
nroot:= -1 * midpt;
IF (X >= 0) THEN
NROOT:= MIDPT;
IF (X < 0) AND (N MOD 2 = 0) THEN
HALT;
end;{nroot}


Continued in my posts below is some output from my program above.

[MTd]

The proof is so hard that after 3 and 1/2 years after the proof was announced, the method is not understood yet!

[numberphile]

We'll get the full 500 pages into the next one maybe?

Basically the Japnese guy proved the conjecture explained throughout this video. The details of his proof is well beyond our powers!

[AdamFerrari]

For those watching in 2023: This proof has STILL not been accepted by the community. So the A B C conjecture is still unsolved 😮