Fermat's Method of Infinite Descent

My second note for TorQue Group - Here we go :) ......

$\textbf{Q.}$Solve in non-negative integers the equation:- $x^{3}+2y^{3} = 4z^{3}$

new

$\textbf{Solution.}$ Clearly we can observe the $(0,0,0)$ is a solution.

Next let us suppose that $(x_1,y_1,z_1)$ be a non-trivial solution.Since $2^{1/3}$ and $4^{1/3}$ are both irrational it is clear that $x_1 > 0,y_1 > 0,z_1 > 0$.

From $x_1^{3}+2y_1^{3} = 4z_1^{3}$,it is quite evident that $x_1$ must be even.So we can write $x_1 = 2x_2$.(where $x_2$ is a positive integer).Then we have $4x_2^{3}+y_1^{3} = 2z_1^{3}$.Hence now we have $y_1=2y_2$.Similar;y in the next step we will have $z_1 = 2z_2$.So finally we can continue this procedure over and over again to obtain a decreasing sequence of integers $x_1 > x_2 > x_3 > ....$.But since $x_1$ is a finite positive integer this sequence surely cannot go on forever with going negative.Hence we have a contradiction and no other solution is possible.

$\textbf{Introduction:}$Fermat's method of infinite descent is essentially the contrapositive of mathematical induction that can be used to disprove statements.In the language of the ladder metaphor, if you know you can’t reach any rung without first reaching a lower rung, and you also know you can’t reach the bottom rung, then you cannot reach any rung.

dominoes falling??no??

$\textbf{Statement:}$ It can be stated as follows:

Let $k$ be a non-negative integer.

Whenever $P(m)$ is true for an integer $m > k$ ,then there must exist some smaller integer $j$ with $m>j>k$ for which $P(j)$ is true.

The $P(n)$ is false for all $n > k$

$\textbf{Intuition:}$That is, if there were an n for which $P(n)$ was true, one could construct a sequence $n > n_1 > n_2 > · · ·$ all of which would be greater than$k$ but for the non-negative integers, no such infinite descending sequence exists.

$\textbf{Variants:}$The FMID has two variants that can be extremely useful especially in the solution of Diophantine equations.They are

$1.$There is no sequence of non-negative integers $n_1 > n_2 > · · ·$ .

$2.$If the sequence of non-negative integers $n_i$ with $1 \le i$ satisfies the inequalities $n_1 ≥ n_2 ≥ · · ·$ , then there exists $i$ such that $n_i = n_{i+1} = · · · .$

We used the variant 1 itself to solve the problem at the beginning of this section.

$\textbf{Q.}$Find the maximal value of $m^{2} + n^{2}$ if m and n are integers between $1$ and $1981$ satisfying $(n^{2} - mn - m^{2})^{2} = 1$. [22nd IMO]

$\textbf{Solution:}$Clearly $(m,n) = (1,1)$ is a trivial solution.Moreover,putting $m=n$ we get $m=n=1$ .

Also if a pair $m,n$ satisfies the relation and $0 < m < n$ then we have $m < n\le 2m$[This can be shown by ptuung the values in the equation].Then completing he square we get : $(n^{2} - mn - m^{2})^{2} = ((n - m)^{2} + mn - 2m^{2})^2$ $= ((n - m)^{2} + m(n - m) - m^{2})^{2}$ $= (m^{2} - m(n - m) - (n - m)^{2})^{2}$

Clearly $m,n-m$ satisfies the same relation.

By Variant 2, the transformation $(m, n) → (n-m,m)$ must terminate after finitely many steps, and it terminates only when $m = n = 1.$.Hence all the pairs satisfying the above relation can be obtained by the inverse transformation $(m, n) → (n,m + n)$ several times:$(1, 1) → (2, 1) → (3, 2) → (5, 3) →· · ·$

So the components of all such pairs are Fibonacci numbers.The largest Fibonacci number less than $1981$ is $F_{16} = 1597$,so the answer to the problem is $F_{15}^{2}+F_{16}^{2} = 3524578$.

Fib

$\textbf{Historical Background:}$The method of infinite descent is associated with the name of Pierre de Fermat because he was probably the first to state it explicitly even though Euclid makes use of the infinite descent in his elements (see problem2).In a letter to Christian Huygens Fermat claimed infinite descent as his own :

"I have finally organized this according to my method and shown that if a given number is not of this nature there will be a smaller number which also is not, then a third less than the second, etc., to infinity, from which one infers that all numbers are of this nature."

Fermat

Here are a few practice problems please post the answers in the comment box :) . To view my note on the partitioning of integers click here

$\textbf{Problem 1.}$Using Fermat's principle of Infinite descent prove that $\sqrt{2}$ is irrational.

$\textbf{Problem 2.}$Prove that composite number has a prime divisor.(Euclid Elements VII.31)

$\textbf{Problem 3.}$Solve in positive integers the equation $x^{2} + y^{2} + x + y + 1 = xyz$.

$\textbf{Problem 4.}$ Solve in positive integers $x, y, u, v$ the system of equations

$x^{2} +1 = uy$, $y^{2} + 1 = vx.$

$\textbf{Problem 5.}$Prove that if there is a triple (x, y, z) of positive integers such that $x^{2} + y^{2} + 1 = xyz$, then z = 3.Find all such triples.

$\textbf{Problem 6.}$Prove that $\sqrt{k}$ is irrational for all k not a perfect square.

$\textbf{Problem 7.}$Find all integers $x, y, z$ satisfying $x^{2} + y^{2} + z^{2} - 2xyz = 0$. [Korean Mathematical Olympiad]

$\textbf{Problem 8.}$Find all solutions of $a^{2} + b^{2} = 3(s^{2} + t^{2})$

My next post will be on stacking problems(which I was supposed to do in this one :P) Good Day!!