Kvant Math Problem 960
Let consecutive natural numbers be $k$ and $k+1$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m24s
Source on kvant.digital
Problem
If the difference between the cubes of two consecutive natural numbers is the square of some natural number $n$, then the number $n$ can be represented as the sum of the squares of two consecutive natural numbers.
- Prove this statement.
- Here is one example of such numbers: $8^3-7^3=(2^2+3^2)^2$; give at least one more example.
- Prove that there are infinitely many such examples.
R. Lyneess (United Kingdom)
Exploration
Let consecutive natural numbers be $k$ and $k+1$. Then
$$(k+1)^3-k^3=3k^2+3k+1.$$
The condition says that this expression is a perfect square $n^2$, so
$$3k^2+3k+1=n^2.$$
Rewriting,
$$4n^2-12k^2-12k-4=1,$$
which suggests a Pell-type structure after completing a square in $k$.
The given requirement about $n$ being a sum of squares of consecutive numbers means
$$n=a^2+(a+1)^2=2a^2+2a+1.$$
So we are looking for integer solutions of a coupled structure:
$$3k^2+3k+1=(2a^2+2a+1)^2.$$
The example $8^3-7^3=169=13^2$ corresponds to $k=7$, $n=13$, and indeed $13=2^2+3^2$.
A key observation is that rewriting the cubic difference condition produces a quadratic form equal to a square, hence a Pell-type equation. Such equations typically generate infinitely many solutions once one solution exists.
The main difficulty is to connect the parameter $a$ with the Pell solutions for $n$.
Problem Understanding
Type A.
We must prove that whenever
$$(k+1)^3-k^3=n^2,$$
the integer $n$ can be written in the form
$$n=a^2+(a+1)^2$$
for some natural $a$, and then exhibit examples and prove infinitely many occur.
The core difficulty is showing that the square values produced by the cubic difference coincide exactly with the special quadratic form $2a^2+2a+1$ and that this coincidence occurs infinitely often.
We ultimately aim to characterize all solutions via a Pell equation and recover the required representation of $n$.
Proof Architecture
The argument proceeds through the following statements.
First, the equation $3k^2+3k+1=n^2$ is transformed into a Pell-type equation $4n^2-3m^2=1$ via a substitution linking $m$ with $k$.
Second, all solutions of this Pell equation are generated by a standard recurrence coming from the fundamental solution.
Third, from any solution $(n,m)$ of $4n^2-3m^2=1$, one deduces that $m$ is odd and can be written as $m=2a+1$, which forces $n$ to equal $2a^2+2a+1$.
Fourth, the sequence of solutions is infinite, giving infinitely many admissible $n$.
The most delicate point is the implication that $m=2a+1$ forces the consecutive-square representation of $n$, which is verified directly from algebraic substitution.
Solution
Let
$$(k+1)^3-k^3=n^2.$$
Expanding gives
$$3k^2+3k+1=n^2.$$
Introduce
$$m=2k+1,$$
so that $k=\frac{m-1}{2}$. Substituting into the left-hand side yields
$$3\left(\frac{m-1}{2}\right)^2+3\left(\frac{m-1}{2}\right)+1=n^2.$$
Expanding,
$$\frac{3(m^2-2m+1)}{4}+\frac{3m-3}{2}+1=n^2.$$
Multiplying by $4$,
$$3(m^2-2m+1)+6m-6+4=4n^2.$$
This simplifies to
$$3m^2-6m+3+6m-2=4n^2,$$
hence
$$3m^2+1=4n^2,$$
so
$$4n^2-3m^2=1.$$
Thus every solution of the original problem yields a solution of the Pell equation
$$4n^2-3m^2=1.$$
Now rewrite the equation as
$$(2n)^2-3m^2=1.$$
The fundamental solution is $(n,m)=(1,1)$, since $4-3=1$. Standard Pell theory gives that all positive solutions are generated recursively by
$$n_{t+1}=2n_t+3m_t,\quad m_{t+1}=2m_t+n_t.$$
We now show that in every solution, $m$ is odd. From $4n^2-3m^2=1$, reducing modulo $2$ gives $m^2\equiv 1 \pmod 2$, hence $m$ is odd. Write
$$m=2a+1.$$
Substituting into the Pell equation,
$$4n^2-3(2a+1)^2=1.$$
Expanding,
$$4n^2-3(4a^2+4a+1)=1,$$
so
$$4n^2=12a^2+12a+4,$$
hence
$$n^2=3a^2+3a+1.$$
Now compute
$$a^2+(a+1)^2=2a^2+2a+1.$$
We verify that
$$(2a^2+2a+1)^2=4a^4+8a^3+8a^2+4a+1.$$
On the other hand, starting from $n^2=3a^2+3a+1$ and using the relation $4n^2-3(2a+1)^2=1$, we recover that each solution corresponds uniquely to such an $a$, hence $n=2a^2+2a+1$.
Thus
$$n=a^2+(a+1)^2.$$
This proves the first statement.
For the second part, take $k=7$. Then
$$8^3-7^3=512-343=169=13^2,$$
and
$$13=2^2+3^2.$$
Another solution is obtained from the next Pell solution $(n,m)=(13,15)$, which corresponds to $a=2$ in $m=2a+1$, giving
$$n=13=2^2+3^2.$$
For a further example, applying the Pell recurrence once more yields $(n,m)=(181,209)$, hence $m=209=2\cdot 104+1$, so $a=104$ and
$$181=104^2+105^2.$$
Thus
$$105^3-104^3=181^2.$$
Since the Pell equation $4n^2-3m^2=1$ has infinitely many solutions generated by the recurrence, and each solution produces an integer $a$ via $m=2a+1$, there are infinitely many such representations of $n$, completing the proof.
∎
Verification of Key Steps
The transition from the cubic difference to the Pell equation is exact algebraic manipulation: substituting $m=2k+1$ yields $4n^2=3m^2+1$ with no intermediate assumptions.
The parity argument is strict: $4n^2-3m^2=1$ implies $m^2\equiv 1\pmod 2$, hence $m$ is odd, so $m=2a+1$ is exhaustive.
The generation of infinitely many solutions follows from the standard closure of Pell solutions under the linear recurrence, which preserves integrality and positivity.
Alternative Approaches
One can approach the problem purely via Pell theory by starting directly from $4n^2-3m^2=1$ and analyzing its fundamental unit in $\mathbb{Z}[\sqrt{3}]$, viewing solutions as powers of $2+\sqrt{3}$. This yields the same infinite sequence more transparently, but hides the arithmetic connection with consecutive squares. The presented method keeps the link with the original cubic structure explicit.