<p>A PROOF OF FERMAT’S LAST THEOREM</p>
<p>The solution to Fermat’s last conjecture is given below:
Fermat’s last theorem states that
a^n+b^n≠c^n
for n, a, b and c belonging to the set of positive integers, and n is greater than 2.
Now, we know that an+bn<(a+b)n for n>=2.
The proof is based on reduction ad absurdum, or proof by contradiction.
Therefore, let
a^n+b^n=c^n
for n>2.
Cn<(a+b)n
Taking the nth root on both sides, we get
c<a+b this=“” clearly=“” gives=“” us=“” a=“” triangle=“” inequality.=“” with=“” any=“” three=“” positive=“” integers=“” in=“” such=“” relation,=“” we=“” can=“” consruct=“” triangle.=“” an=“” extension=“” to=“” further=“” affirm=“” the=“” mathematical=“” validity=“” of=“” my=“” statements:=“” cn=“an+bn” an+bn<(a+b)n=“” for=“” n=“”>=2 by binomial theorem
As a result, cn<(a+b)n
Taking the nth root on both sides, c<a+b
To prove the statement that this is a triangular inequality, we have to affirm that a<c+b, and b<a+c.
an= cn-bn
cn-bn<(c+b)n
As a result, an<(c+b)n
Taking the nth root on both sides, we get
a<c+b
We can repeat the same steps to arrive upon the relation that b<a+c.</a+b></p>
<p>Now, it is known, according to the cosine rule of a triangle, that
c^2=a^2+b^2-2ab<em>cosC…………(i)
Thus, keeping this relation in mind, cn can be written as 〖〖[c〗^2]〗^(n/2), which can again be written as
[a^2+b^2-2ab</em>cosC]n/2 , on substitution from (i).
We know that c^n is an integer. Thus, [a^2+b^2-2ab<em>cosC]n/2 should be an integer. It is known that ‘a’ and ‘b’ are integers. However, |cos〖C|<1〗. This is because C≠0⁰, which makes cos C=1. C cannot be equal to 0⁰ because it is known that a triangle is formed by the three integers a, b and c, and no angle in a triangle is 0⁰.
Hence, unless C= 90⁰, in which case cos C=0, 2ab</em>cosC will be a non-integer, as an integer times a non-integer is a non-integer. Subsequently, a^2+b^2-2ab<em>cosC will also be a non-integer, as an integer (a^2+b^2 is an integer) minus a non-integer is a non-integer.
It is also known that a non-integer raised to any rational power is a non-integer. Hence, [a^2+b^2-2ab</em>cosC]n/2 will also be a non-integer. This contradicts the assumption that cn is an integer. Hence, a contradiction is reached.
If cos C is taken to be equal to 0, then we have cn= 〖(a^2+b^2)〗^(n/2). However, the binomial expansion of 〖(a^2+b^2)〗^(n/2) will yield a sum that is greater than (a^n+b^n). Hence, the only suitable choice of n is 2, which leaves us with c^2=a^2+b^2 as the only possibility. Pythagorean triplets are the only solutions to Fermat’s Last Conjecture.</p>