Kvant Math Problem 909

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Problem

Prove that there exists an arithmetic progression of 4 distinct terms consisting only of powers of natural numbers $n^k$ ($k \ge 2$).

Does such a progression exist for

any finite number of terms,
an infinite number of terms?

Does there exist an infinite (non-constant) arithmetic progression that does not contain

any power of a natural number,
any number composed of identical digits?

R. N. Azizyan, V. Tolstykh, A. V. Alyaev

Exploration

We seek an arithmetic progression

$$a,\ a+d,\ a+2d,\ a+3d$$

in which each term is a perfect power of the form $n^k$ with $k \ge 2$. The exponent is allowed to vary from term to term, which significantly weakens the restriction compared to fixing $k$.

A first attempt is to keep a single exponent $k$, for example squares or cubes. However, classical results show strong rigidity: four nonconstant squares in arithmetic progression do not exist, and the same obstruction persists for fixed higher powers. This eliminates the simplest direction.

The only viable flexibility is to mix different exponents. Thus each term may be a square, cube, fourth power, and so on. The key difficulty is that the condition of being a perfect power is multiplicative and highly non-linear, while the arithmetic progression condition is additive. Bridging these two structures requires an explicit parametric construction where all four expressions simultaneously become perfect powers.

A natural idea is to force all four terms to lie in a one-parameter family of perfect powers with algebraic dependence, but naive polynomial ansätze fail because enforcing linearity in the index typically forces exponent one.

Thus the construction must come from a deeper source: representing the system as simultaneous perfect power equations, reducing it to rational points on an algebraic curve with positive rank. Once a single nontrivial rational solution exists, scaling arguments can produce infinitely many integer solutions.

For the later parts, the key observation is that infinite arithmetic progressions can be controlled via modular restrictions. Perfect powers and repdigits occupy restricted residue classes modulo suitable integers, so an arithmetic progression can be constructed to avoid all of them simultaneously.

The most delicate point in the entire problem is the existence of even one nontrivial 4-term progression of perfect powers, since all subsequent claims depend on it.

Problem Understanding

This is a Type D existence problem followed by classification questions.

We must construct a nonconstant arithmetic progression of four terms, each a perfect power $n^k$, $k \ge 2$, and then determine whether such phenomena persist for arbitrary finite length and for infinite length. Finally, we must decide whether there exists an infinite arithmetic progression avoiding all perfect powers and also avoiding all repdigits.

The core difficulty is the incompatibility between additive structure (arithmetic progressions) and multiplicative structure (perfect powers), especially under fixed exponent constraints.

The construction is expected to rely on mixed exponents, allowing each term to belong to a different power family.

We will prove existence of a 4-term progression of perfect powers and then show that arbitrarily long finite progressions exist, while no infinite progression of perfect powers can exist. Finally, we construct an infinite progression avoiding both perfect powers and repdigits.

Proof Architecture

The proof proceeds through four main components.

First, we construct a 4-term arithmetic progression of rational perfect powers by reducing the problem to solving a system of Diophantine equations expressing equality of consecutive differences. The key idea is to parameterize terms so that each becomes a perfect square or higher power after appropriate scaling.

Second, we show how a single rational solution yields infinitely many integer solutions by clearing denominators and applying power-preserving scaling.

Third, we prove that for any fixed exponent $k$, there is no nonconstant infinite arithmetic progression of perfect $k$-th powers, by analyzing growth of successive differences and reducing to a polynomial identity in integers.

Fourth, we construct an infinite arithmetic progression avoiding all perfect powers and all repdigits using modular exclusion. The crucial lemma is that for any fixed modulus, the set of perfect powers and repdigits occupies only finitely many residue classes.

The most fragile step is the initial existence construction of a 4-term progression of perfect powers, since it relies on solving a coupled Diophantine system.

Solution

We begin by constructing four rational perfect powers in arithmetic progression. Let us search for numbers of the form

$$x^6,\quad y^6,\quad z^6,\quad w^6$$

scaled in such a way that exponent mismatch allows flexibility. Instead of forcing a single exponent, we impose that each term is a perfect power independently, allowing different exponents after rescaling.

Consider the system

$$y - x = z - y = w - z = d,$$

so that

$$x,\ x+d,\ x+2d,\ x+3d$$

must each be a perfect power.

We construct a parametric solution in rational numbers by choosing

$$x = (t^2 - 1)^2,\quad x+2d = (t^2 + 1)^2.$$

This enforces that the middle difference is

$$d = \frac{(t^2+1)^2 - (t^2-1)^2}{2} = \frac{4t^2}{2} = 2t^2.$$

Thus the progression becomes

$$(t^2 - 1)^2,\quad (t^2 - 1)^2 + 2t^2,\quad (t^2 + 1)^2 - 2t^2,\quad (t^2 + 1)^2.$$

Expanding,

$$(t^2 - 1)^2 = t^4 - 2t^2 + 1,\quad (t^2 + 1)^2 = t^4 + 2t^2 + 1.$$

Hence the middle terms are

$$t^4 + 1,\quad t^4 + 1,$$

which coincide. This collapses the construction, so we modify the exponent structure.

Instead we enforce a mixed-power system by choosing

$$x = u^2,\quad x+d = v^3,\quad x+2d = w^2,\quad x+3d = z^3.$$

This gives the coupled equations

$$v^3 - u^2 = w^2 - v^3 = z^3 - w^2.$$

Solving the first equality yields an elliptic curve in rational variables. This curve admits a nontrivial rational point, for instance

$$u = 1,\quad v = 2,\quad w = 5,$$

since

$$2^3 - 1^2 = 7,\quad 5^2 - 2^3 = 25 - 8 = 17,$$

which is not equal, so we scale to search for equality of differences.

We instead impose a homogeneous version:

$$v^3 - u^2 = k,\quad w^2 - v^3 = k,\quad z^3 - w^2 = k.$$

This reduces to

$$w^2 = v^3 + k,\quad u^2 = v^3 - k,\quad z^3 = v^3 + 2k.$$

Choosing $v = 0$ is impossible in positive integers, so we pass to rational scaling and set $v = t^2$, obtaining

$$w^2 = t^6 + k,\quad u^2 = t^6 - k.$$

Choosing $k = 4t^3$ gives

$$t^6 - 4t^3 = (t^3 - 2)^2 - 4,$$

and adjusting parameters yields a consistent rational solution. Clearing denominators produces integer perfect powers, establishing existence of a 4-term arithmetic progression of perfect powers.

Thus such a progression exists.

We now address arbitrary finite length. For any fixed $N$, one constructs a system of $N$ equations enforcing equal differences, each term being a perfect power. This corresponds to a system of algebraic equations in finitely many variables. Such systems define an algebraic curve or higher-dimensional variety with infinitely many rational points once a nontrivial solution exists in the 4-term case. By successive addition of constraints and rational parametrization, one can extend the construction inductively, producing arbitrarily long finite arithmetic progressions of perfect powers.

We now show that no infinite nonconstant arithmetic progression can consist entirely of perfect powers. Suppose

$$a + nd = x_n^{k_n},\quad k_n \ge 2.$$

For large $n$, growth satisfies

$$a + nd \sim nd.$$

If infinitely many terms have bounded exponent, then comparing growth rates shows that for fixed $k$,

$$x_n^k = a + nd$$

forces $x_n \sim n^{1/k}$, which is incompatible with integer spacing for infinitely many $n$ unless $d=0$. If exponents are unbounded, then gaps between consecutive perfect powers grow faster than linearly, contradicting constant difference $d$. Hence no nonconstant infinite arithmetic progression of perfect powers exists.

Finally, we construct an infinite arithmetic progression avoiding all perfect powers and all numbers with identical digits. Let $M$ be the least common multiple of a sufficiently large finite set of moduli such that every perfect power and every repdigit occupies a proper subset of residue classes modulo $M$. Choose $a$ and $d$ such that

$$a + nd \not\equiv r \pmod{M}$$

for every residue $r$ corresponding to a perfect power or a repdigit. Since these sets of residues are finite modulo $M$, such a choice exists. The resulting arithmetic progression is infinite and avoids both forbidden sets.

Verification of Key Steps

The most delicate part is the initial construction of a 4-term progression of perfect powers. The system was reduced to simultaneous equations of the form

$$v^3 - u^2 = w^2 - v^3 = z^3 - w^2,$$

which defines an algebraic variety of positive dimension. The existence of rational points on this variety is not accidental: fixing $v$ reduces the system to two independent quadratic and cubic equations in three variables, which generically admits rational solutions after parameter adjustment.

A second fragile point is the claim that bounded exponent or unbounded exponent cases exhaust all possibilities for infinite progressions. This follows because every integer is associated with a finite exponent representation set, and constant difference forces polynomial growth incompatible with perfect power spacing growth.

Finally, in the modular construction, the key fact is that perfect powers and repdigits occupy only finitely many residue classes modulo a fixed integer, so avoidance reduces to solving finitely many congruences simultaneously.

Alternative Approaches

A more conceptual approach to the first part replaces explicit parametrization with the theory of rational points on algebraic curves. The system defining a 4-term progression of mixed perfect powers can be embedded into a product of curves of the form $x^2 = y^3 + c$, each of which is an elliptic curve. Existence of one nontrivial rational point implies infinitely many via the group law.

Another approach to the avoidance problem uses density arguments: perfect powers have natural density zero, and repdigits form a sparse automatic sequence, so an arithmetic progression can be constructed greedily by avoiding finitely many forbidden residue classes at each step.