A simple solution.
French mathematician Pierre de Fermat wrote a statement around 1.637 A.D. that said “...no three positive integers a, b and c satisfy the equation aⁿ + bⁿ = cⁿ for any integer value of n greater than 2”.
The
exceptions are the trivial solutions (0,
1, 1), (1, 0, 1) and (0, 0, 0), of course. The cases n=1 and n=2 have been
known since antiquity to have infinitely many solutions (Wikipedia dixit).
Fermat
didn’t write proof of his statement because, apparently, there was not enough blank space left in the page to do it. Wait, what?
In 1995
(more than 350 years later) mathematician Andrew Wiles (with the help of his
colleague Richard Taylor) proved Fermat’s statement in a published article… The first complete, valid proof of that
statement ever! Because of that great achievement, en 2016 Wiles
was awarded with the Abel Prize (an award created in Sweden to compensate for
the lack of a “Nobel of Mathematics”). Nowadays
that statement, its proof and implications are known as the Last Theorem of
Fermat-Wiles. And Maths are better
because of it.
See? A happy ending.
When I
read that story, my intuition was that Fermat –genius as he was- didn’t think
of something extraordinarily complicated or have an astounding revelation, but
that he thought of something really simple…
And that was the reason why he didn’t bother to write any proof about
his statement anywhere else.
In my
country –Spain- there’s a saying: “Ignorance is so much daring”. Why do I tell you this? Because a thrilling idea crossed my head:
“Well, if it’s got a really simple solution, maybe I will be able to find it,
too”.
For the
last three weeks, I’ve been thinking about it, notebook and pen in my hands... And now I
have a simple solution to Fermat’s statement.
And by “simple” I mean it: it’s such a really simple
solution that I´m sure many, many, many people must have thought and found it
by themselves before I did. So this must be something already KNOWN, right? And maybe they teach it in the first academic
year of Maths at high school...
But then, why was it never accepted as orthodox proof?
I don’t know. My guess is, when 17th century
mathematicians started trying to solve Fermat’s statement, they soon found that
it may had implications over other aspects of Geometry and Arithmetic; and those
implications would have some other implications… and everything quickly turned into
something with a lot of complexity; more and more of it as time went by.
Pierre de Fermat was a genius; and I am under the impression that right before writing his statement, proof for it came to his
mind in a whole… or in a matter of
minutes, at most. Maybe he already knew it
beforehand.
I’m not a genius: I’m only good old Carlos… With my old training –from the eighties- to
be a Primary School teacher and my EGB professor’s diploma. And my IQ is the one of an average person (the
years of my life have proved that once and again).
But this is my blog and this post, the result of my curiosity and my hard work. What the heck, I was eager to tell someone.
For the rest of the post, I’m going to refer to Fermat’s statement’s expression as aⁿ = bⁿ + cⁿ (for what all those numbers can’t satisfy) or aⁿ ≠ bⁿ + cⁿ (for what they end up doing).
There’s something else. Something important, in fact.
Fermat studied numbers, all right. But it’s quite clear he studied them in the guise of the measures of the sides of triangles. Think about it: Fermat’s statement is kind of an evolution of Pythagoras’ theorem… And the numbers he talks about are positive integers, simple but ideal for the length of triangle sides.
So thinking a lot about triangles (and I mean all possible triangles whose sides are the length of positive integers) is the way to solve Fermat’s statement, don’t you agree?
I guess it won’t be a surprise if we start with…
CASE 1: RIGHT-ANGLED
TRIANGLES.
Three
sides, a, b, c: all of them the
length of positive integers (or natural numbers).
a for the hypotenuse; b, c for
the legs or catheti; a > b, a > c.
The
triangle and its numbers satisfy the parity of Pythagoras’ theorem: a² = b² + c².
We
calculate a new parity multiplying both sides by a…
a · a² = a · (b² + c²) → a³ = ab² + ac²
Since a > b, then ab² > b³; since a > c,
then ac² > c³.
Look: (ab² + ac²) > (b³ + c³) → a³ > (b³ +
c³) → a³ ≠ b³ + c³.
And the
same reasoning applies to every value of n from 4 (included) to infinite… So every right triangle verifies that aⁿ ≠ bⁿ + cⁿ.
CASE 2: SCALENE TRIANGLES.
Three
different sides, a > b > c:
all of them positive integers (or natural numbers).
I’ve
drawn an external height, y. To do that, I showed the straight line cathetus c lies on by drawing an extension, x. Both x and y are positive real numbers (if natural, rational o irrational is
not relevant).
b²= x² + y²
a² = (c + x)² + y²
Let’s
start with the one below.
a²
= (c + x)² + y² → a² = c² + x² + 2cx + y² →
→
a² = c² + 2cx + x² + y² → a² = c² + 2cx + (x² + y²) →
…and
then, we substitute that last parenthesis…
→ a² = c² + 2cx + b² →
→ a² = b² + c² + 2cx
We’re
going to proceed exactly like we did in CASE 1: First, we multiply both members
in the parity by a…
a
· a² = a · (b² + c²
+2cx) →
→ a³ = ab² + ac² + 2acx
Since a > b
→ ab² > b³; since a > c → ac²
> c³; and 2acx is a positive
real number. So…
ab² + ac² + 2acx > b³ +
c³ →
→ a³ > b³ + c³.
And the
same reasoning applies to every value of
n from 4 (included) to infinite: the real value of aⁿ is more complex and always
will be bigger than whatever results from
bⁿ + cⁿ… Exactly, aⁿ ≠ bⁿ + cⁿ.
---
If you
allow me… Many rectangle triangles are
scalene triangles, but not all of them.
Of course, they all verify aⁿ ≠
bⁿ + cⁿ.
CASE 3: ISOSCELES
TRIANGLES.
Case 3.1: b = c, a <
b, a < c.
Since a < b → a³ < b³; since a < c → a³ < c³; and considering
both truths together… a³ < b³ + c³.
By the
same reasoning, a < b → aⁿ < bⁿ;
a < c → aⁿ < cⁿ.
And
then…
aⁿ < bⁿ + cⁿ → aⁿ ≠ bⁿ + cⁿ.
Case 3.2: a = b, c ≠ a, c ≠ b; and remember, a > 0, b > 0, c > 0 (because any side of any triangle
measures more than zero). Then, if we try…
aⁿ = bⁿ + cⁿ →
→ aⁿ – bⁿ = cⁿ → 0 = cⁿ
…Which
is impossible. So aⁿ
≠ bⁿ + cⁿ.
Case 3.3: b = c, a > b, a > c.
b² = x² + y²
a² = (c + x)² + y²
Let’s expand the second expression…
a² = c² + x² + 2cx + y²
→
→ a² = c² + 2cx + x² +
y² →
→ a² = c² + 2cx + (x² +
y²)
And we
substitute what’s inside the parenthesis as told by the first expression up
there:
a² = c² + 2cx + b² →
→ a² = c² + b² + 2cx →
→ a² = b² + c² + 2cx
We multiply both sides of the parity by
a…
a · a² = a · (b² + c² +
2cx) →
→ a³ = ab² + ac² + 2acx.
Since a > b → ab² > b³; a > c → ac² > c³… And 2acx
is a positive real number (which only adds here)…
ab² + ac² + 2acx > b³ +
c³ →
→ a³ > b³ + c³ → a³ ≠
b³ + c³.
And the
same reasoning applies to every value of
n from 4 (included) to infinite: the real value of aⁿ is more complex and always
will be bigger than whatever results from
bⁿ + cⁿ…
Exactly, aⁿ ≠ bⁿ + cⁿ.
---
Just a
curiosity: there are rectangle triangles which are isosceles triangles,
too. Their same-length sides are the
legs or catheti. The remaining side (the
different, bigger one) is the hypotenuse.
And yes,
they all verify aⁿ ≠ bⁿ + cⁿ.
CASE 4: EQUILATERAL
TRIANGLES.
Three
sides, a = b = c. Right, they’re all the same length.
If we
try aⁿ = bⁿ + cⁿ → aⁿ = aⁿ + aⁿ → aⁿ =
2aⁿ.
Zero looks
like a solution for that parity, but not one we are eager to hold to… Because any side of a triangle must be bigger
than zero. We’re looking for positive
integers, remember?
In this
case we contemplate (the one where a = b
= c and a > 0, b > 0, c > 0),
any positive integers will give us aⁿ ≠
bⁿ + cⁿ.
That’s
what any and all equilateral triangles will verify.
CONCLUSION:
We just proved all possible triangles whose sides are the measure of positive integers verify the expression aⁿ ≠ bⁿ + cⁿ for values of n > 2.
Which, if I’m not mistaken, pretty much equals the meaning of Fermat’s statement (yes, the one in the first paragraph of this blog post). And we did it in a quite simple way. A simple solution, I told you.
Oh, no. I'm sorry. It's me, Carlos, some days later...! I just realized there's something else to prove if I want to get to a complete & solid simple solution to Fermat's statement. You can find it in this brand new other blog post clicking here).
Thank you for reading.
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