# This note has been used to help create the Fermat's Method of Infinite Descent wiki

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!! 6 years ago

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Such well explanation, wonderful!

- 6 years ago

Thanks!!Don't forget to check out my note on Stacking problems

- 6 years ago

Are u majoring in maths or something. Because i never faced these interesting theorems and conjectures during my high school and not even in graduation. They are not in our courses.

- 5 years, 8 months ago

Can I replace the method of infinite descent with a contradiction proof as follows:

Suppose that we have the smallest non-trivial solution. Using one step of the method of infinite descent, we arrived at a smaller solution, contradiction. Therefore there are no non-trivial solutions.

This method seems to be much simpler than the entire infinite descent method... Is there any advantage the method of infinite descent has over this way?

- 5 years, 12 months ago

I actually prefer your way over the infinite descent method. Looking at the smallest (and largest cases) sometimes gives us a lot of useful information about a problem. This technique of checking the extreme cases is called the extreme principle and it is one of my most favorite proof techniques.

- 5 years, 11 months ago

Sometimes finding and proving theminimum possible value can be tedious....there infinite descent is much more effective I thnk...check out Eulrers proof of fermats two square theorem

- 5 years, 11 months ago

What yo said is essentially variant 1 of infinite descent....Is it not???

- 5 years, 12 months ago

They are two forms of the same thing not two different steps....Each applied in different cases....

- 5 years, 12 months ago

Really Appreciable Notes

- 5 years, 5 months ago

very nice, can you repost the first torque note please?

- 6 years ago

I have put the link just below the picture of Fermat and above the problems :)

- 6 years ago

For example 2, you mentioned that "the above relation can be obtained by transformation $(m,n)$ into $(n, m+n)$.

May I know how did u transform and why did we have to transform? Thanks!

- 6 years ago

First we obtained a trivial solution which is $(m,n) = (1,1)$.The we proved that whenever $(m,n)$ is a solution $(n-m,m)$ is also a solution and hence this decreasing sequence cannot go on forever and must terminate only when m = n = 1. So from (1,1) we perform the reverse transformation that is from $m,n) -> (n,m+n)$ to obtain the other pairs of solutions and getting the Fib sequence in return......... Note that the sequence will terminate iff m = n = 1 which follows from variant 2.....If you have any doubt feel free to ask....

- 6 years ago

I was very fascinated to read this note I am not familiar with transformations could u explain it in more simple terms.

- 4 years, 2 months ago

Could anyone give a hint for Problem 1?

- 6 years ago

Suppose $\sqrt{2}$ is rational.

Then $\sqrt{2}$ can be written as $\sqrt{2} = \frac{a}{b}$.

So we get $2a^{2} = b^{2}$.

Since $b$ must be even,

so.........................D.I.Y...............................

- 5 years, 11 months ago

Thank you! :)

- 5 years, 10 months ago

As for problem one can approach it the conventional way but instead of taking for granted that $gcd(p,q) = 1$ there try to use infinite descent(The general idea)....It's the easiest problem in the section....

- 6 years ago

Suppose the contrary and let $\sqrt{2}=\dfrac{p}{q}$ for relatively prime positive integers $p,q$.

- 5 years, 12 months ago

Can anyone post hints to Problem 3 onwards??? Or maybe post 1 solution? Thanks! :)

- 5 years, 10 months ago

God I love your notes. So awesome.

- 5 years, 8 months ago