hello everyone I'm going to do some
proper maths today I'm going to talk
about this equation x the n + y the n =
z to the N I want to find Solutions
where X Y and Zed are integers so let's
take some examples let's do n = 1 then
that's quite easy really I've got 2 + 3
= 5 would be an integer solution I hope
I haven't blown your mind with maths too
early if we took n = 2 we could have 32
+ 42 = s which is called a Pythagorean
triple for Nal 3 there are no integer
Solutions and that's the same for n four
or five or six and that's because this
is fermat Last Theorem which was proven
by Andrew WS which says that there are
no integer Solutions when N is a
positive integer greater than two but
the thing about fermatas theorem is it
says nothing about when the exponent is
not a positive integer what about if it
was a negative number or a rational
number or an irrational number can we
find integer Solutions then well we can
here's an example 32 to ^5 + 243 ^5 =
3,125 to ^ 1/5 or here's another example
27 ^ 2/3 + 64 to^ 2/3 = 125 to^ 2/3 but
some of you may have already noticed
that these are actually cheats all I've
done here is taken the two examples I've
already shown you and hidden them in a
slightly clever way and it turns out
that when the exponent is rational these
kind of cheats are all the integer
solutions that we can find there are no
other Solutions there is a truly
marvelous proof of this which this video
is too short to contain so let me give
you the gist what we're trying to solve
is this equation so we're having the
exponent be a rational number which I've
written as k k/ M here or to put it
another way we are taking the M roots of
x^ k y^ k and z to the power K so we're
looking for these nth Roots now they are
also algebraic numbers so when you're
taking real Roots we can use properties
of algebraic numbers to prove that these
n of the roots are integers and if
they're integers then X De k y to the k
z to the K well they have to be powers
of K obviously and they're going to have
to be poers of M as well so this
equation actually turns into this and
after taking the nth root we get a to ^
K plus b to the^ K = C to the power K A
B and C are integers here which is Fat's
Last Theorem which only has integer
Solutions when K is equal to 1 or two so
we have Solutions but we're restricted
by firmat Last Theorem which seems
perfectly reasonable to me one word of
warning though because there is a sense
where this is true 1 ^ 56 plus 1 ^ 56 =
1 to the^ 56 but that seems like bunk
because clearly the left hand side is
equal to two and the right hand side is
equal to one except it doesn't have to
be for example a 5 6th root of 1 could
be e to Pi I over 3 and another 56 root
of 1 could be e to 5 Pi I over 3 and
third 56 root of one could be one itself
and then this becomes a true equation
and in this case the MTH roots are now
complex Roots so if we allow complex
Roots we can get some strange integer
Solutions but we can prove that only
happens when x = yal z and M is
divisible by 6 which I think is a bit of
a surprise now the proof of that result
uses gawa Theory and if you're
interested in the full proofs of these
results
well I will put a link in the
description incidentally whenever we
have an integer solution with a positive
exponent n we automatically have an
integer solution with a negative n so if
we take X the n + y n = z the n and we
divide through by x * y * Z to the^ n we
will get an integer solution with a
negative exponent so if I took an
example like 3^2 + 42 = 5^2 that will
correspond with 20 ^ - 2 + 15^ - 2 = 12
^ - 2 and this works in both directions
well that was for rational exponents
what about irrational exponents well
first of all do we even have integer
Solutions with an irrational exponent
well yes we do and that's actually quite
easy to show for example we could look
at this 4 squ + 5^ 2 is greater than 6 2
but 4 cubed + 5 cubed is less than 6
cubed so there must be a value for the
exponent n somewhere between 2 and 3
where 4 to the n + 5 the N equal 6 the N
now we know that can't be rational it
must be irrational and in fact if you go
searching for it you'll find that n is
equal to
24879 something something so integer
Solutions with irrational exponents
exist but it would be nice to give you
an example where we don't have to go
searching for the exponent so how about
this let's look at this equation x the n
+ x the n = z the N well if we divide
through now by X the N we would have 2 =
Z /x to the power n and then n is the
logarithm to the base Z /x of two so if
we take an actual example of this let's
take 4 the n + 4 n = 5 the N then we can
solve this for n n would be the
logarithm to the base 54 of 2 which is
equal to
3.16 and some stuff and now let me do
just one last example of an integer
solution to Fat Last Theorem with an
irrational exponent and this one I like
because I think it's quite clever it
uses the golden ratio the golden ratio
which we write as 5 is equal to 1 + < TK
5/ 2 well the golden ratio is the
solution to this equation 1 + 5 = 5^ 2
now let 5al 2^ n then if we do that 1
becomes 1 ^ n 5 becomes 2 to the n and
5^ 2 becomes 4 to the^ n so we've got a
firat Last Theorem type equation here
and you can solve this for n the
exponent will be the logarithm base 2 of
the golden ratio 1 +un 5 over two and
that is equal to 0. 6942 and some stuff
now there are no General results for
firat Last Theorem with irrational
exponents except we know that there are
infinitely many integer Solutions out
there so you might like to try to find
one for yourself and if you have been
thanks for watching