Combinatorics problems that can be solved via infinite descent I'm looking for high school problems that can be solved with the method of infinite descent. Usually, those problems are from number theory, but I would be very happy if someone could provide a problem(s) from combinatorics and/or any other field of mathematics. Here are some problems from number theory:

Prove that a following equations have no nontrivial solutions in $\mathbb{Z}$:

$a^3+2b^3 = 4c^3$

$2a^2+3b^2 = c^2+6d^2$

$x^2 + y^2 + z^2 = 2xyz$

$x^4+y^4 = z^2$
There is no regular pentagon whose vertices are on the lattice points.



Suppose that $ABCDE$ is a regular pentagon where $A,B,C,D,E$ are on the lattice points. Then, let $A',B',C',D',E'$ be the intersection point of $BD$ with $CE$, $AD$ with $CE$, $AD$ with $BE$, $AC$ with $BE$, $BD$ with $AC$ respectively.

Then, since $ABA'E$ is a parallelogram, we see that $A'$ is also on the lattice point. Similarly, we see that $B',C',D',E'$ are on the lattice points.

So, we see that $A'B'C'D'E'$ is a smalller regular pentagon whose vertices are on the lattice points.
The Fundamental Theorem of Arithmetic

Existence

The following can be proven using the Principle of Infinite Descent - see this link.

Proposition 1: Every integer $n \ge 2$ can be factored into primes:
There exists a nonempty family $(p_i)_{\, 1 i \le i \le u }$ of primes such that

$$\tag 1 n = p_1 p_2 \dots p_u$$

Uniqueness

Proposition 2: If an integer $n \ge 2$ is written as

$$\tag 2 n = p_1 p_2 \dots p_u$$

and 

$$\tag 3 n = q_1 q_2 \dots q_v$$

with two nonempty families of primes, $(p_i)_{\, 1 i \le i \le u }$ and $(q_i)_{\, 1 i \le i \le v }$,
then $u = v$ and the families can be brought into a $1:1$ correspondence.

Proof
To get a contradiction, assume the proposition is false. 

An easy factoring argument presents us with an integer $n \ge 2$ satisfying $\text{(2)}$ and $\text{(3)}$ with $p_i \ne q_j$ for any subscripts. Also, both $u$ and $v$ must be greater than $1$ and we can insist that $p_1 \lt q_1$.

The number 

$$\tag 4 n' = (q_1 - p_1) q_2 \dots q_v \ge 2$$

is divisible by $p_1$. If $q_1 - p_1 \ge 2$, by proposition 1 we can factor it into primes and it is impossible for $p_1$ to appear in this factorization. Since we also have $p_1 \ne q_i$ for $i \ge 2$, the integer $n' \lt n$ can now also be presented as an integer with two different prime factorizations.

So by the Principle of Infinite Descent, the proposition is true. $\quad \blacksquare$

Combined Statement

Theorem 3: Every integer greater or equal to $2$ has one and only one
(up to a rearrangement) prime factorization.