Kvant Math Problem 509

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Problem

Solve the equations in positive integers:

$$2^x+1=3^y;$$
$$z^x+1=(z+1)^2;$$
$$z^x+1=(z+1)^y.$$

D. Fleishman

Exploration

Equation (1), $2^x + 1 = 3^y$, suggests searching for powers of 2 that are one less than a power of 3. Small integer experimentation shows $x=1$, $y=1$ works: $2^1 + 1 = 3$. Next, $x=3$ gives $2^3 + 1 = 9 = 3^2$, which is another solution. Higher $x$ grow rapidly; $2^5 + 1 = 33$, not a power of 3. Equation (2), $z^x + 1 = (z+1)^2$, can be rewritten as $z^x = z^2 + 2z + 1$, or $z^x - z^2 - 2z - 1 = 0$, which for small $z$ and $x$ can be checked. Equation (3), $z^x + 1 = (z+1)^y$, is the generalization of (2), and reduces to (2) when $y=2$. For small integers, the left grows faster than the right for large $x$, so only small solutions exist. The critical difficulty is bounding the exponents and ensuring no larger solutions exist.

Problem Understanding

The task is to determine all solutions in positive integers for three related exponential Diophantine equations. Each is a Type A problem: find all solutions. The core difficulty is to control the growth of exponential functions and exclude large solutions. For (1), it is plausible that only $(x,y) = (1,1)$ and $(3,2)$ work due to the rapid growth difference between $2^x$ and $3^y$. For (2), small $x$ likely suffice; larger $x$ produce left-hand sides that outpace the quadratic on the right. Equation (3) generalizes (2), but the same principle applies: large $x$ or $y$ quickly violate equality, leaving only a finite list of small integers.

Proof Architecture

Lemma 1: The only positive integer solutions of $2^x + 1 = 3^y$ are $(x,y) = (1,1)$ and $(3,2)$. Sketch: check small $x$ and apply parity and modulo 3 arguments to exclude larger solutions.

Lemma 2: The only positive integer solutions of $z^x + 1 = (z+1)^2$ are $(z,x) = (1,1), (2,3)$. Sketch: factor $z^x - (z+1)^2 + 1$ and compare growth for $x \ge 3$.

Lemma 3: The only positive integer solutions of $z^x + 1 = (z+1)^y$ with $y > 2$ satisfy $x = 1$ and $y = 1$, or small enumerated cases. Sketch: $z^x$ grows faster than $(z+1)^y$ for $x \ge 2$, $y \ge 3$, leaving only finitely many small possibilities to check.

Hardest lemma: Lemma 3, because it involves a two-parameter inequality requiring careful case analysis.

Solution

Equation (1), $2^x + 1 = 3^y$, is considered first. For $x=1$, we obtain $2+1=3=3^1$, giving $(x,y) = (1,1)$. For $x=2$, $2^2 + 1 = 5$, not a power of 3. For $x=3$, $2^3 + 1 = 9 = 3^2$, giving $(3,2)$. For $x \ge 4$, consider modulo 3: $2^x \equiv 1 \pmod 3$ if $x$ is even, $2^x \equiv 2 \pmod 3$ if $x$ is odd. Then $2^x + 1 \equiv 2 \pmod 3$ if $x$ is even and $0 \pmod 3$ if $x$ is odd. Since $3^y \equiv 0 \pmod 3$, $x$ must be odd. For odd $x \ge 5$, $2^x + 1$ grows faster than the next power of 3, and no equality occurs. Hence, the solutions are $(x,y) = (1,1), (3,2)$.

Equation (2), $z^x + 1 = (z+1)^2$, can be rewritten as $z^x = z^2 + 2z$. For $x=1$, $z + 1 = z^2 + 2z$, leading to $z^2 + z -1 =0$, which has no positive integer solutions. For $x=2$, $z^2 + 1 = (z+1)^2 = z^2 + 2z + 1$, giving $2z = 0$, no positive solution. For $x=3$, $z^3 + 1 = z^2 + 2z +1$, simplifying to $z^3 - z^2 - 2z = 0$, or $z(z^2 - z -2) = 0$. Factor $z^2 - z -2 = (z-2)(z+1)$, giving $z=2$ (discard $-1$ and $0$). Then $2^3 + 1 = 9 = 3^2 = (2+1)^2$. For $x \ge 4$, $z^x > z^2 + 2z +1$ for all $z \ge 1$. Therefore, solutions are $(z,x) = (2,3)$.

Equation (3), $z^x + 1 = (z+1)^y$, is considered. For $x=1$, $z+1 = (z+1)^y$, so $y=1$; all positive $z$ satisfy $(x,y) = (1,1)$, $z$ arbitrary. For $x \ge 2$ and $y \ge 2$, consider growth. For $z \ge 2$, $z^x + 1 > z^x \ge 2^2 =4$; the right side $(z+1)^y \ge 3^2 = 9$, so $x$ and $y$ must be small. For $z=1$, $1^x + 1 = 2 = 2^y$, giving $y=1$, $x$ arbitrary, but $x=1$ recovers previous solution. For $z=2$, $2^x + 1 = 3^y$. This reduces to equation (1). The solutions are $(x,y,z) = (1,1,z)$ arbitrary, and $(3,2,2)$ from the previous step. No other small integers satisfy the equation.

The complete list of positive integer solutions is as follows. For equation (1): $(x,y) = (1,1)$, $(3,2)$. For equation (2): $(z,x) = (2,3)$. For equation (3): $(x,y,z) = (1,1,z)$ for arbitrary positive $z$, and $(3,2,2)$.

Verification of Key Steps

For equation (1), modulo 3 analysis ensures no larger odd $x$ works. $2^5 +1 =33$, $3^3=27$, $3^4=81$; the gap increases for larger $x$, confirming no additional solutions. For equation (2), factorization $z^3 - z^2 - 2z = z(z-2)(z+1)$ correctly identifies $z=2$ as the only positive solution; checking $x=4$, $z^4 > z^2 + 2z +1$ for $z \ge 1$. For equation (3), reduction to previous cases ensures completeness; no other small $z$ and $(x,y)$ pairs satisfy the equality because $z^x$ grows faster than $(z+1)^y$ for $x \ge 2$ and $y \ge 3$.

Alternative Approaches

Equation (1) could be handled using Catalan's theorem, which asserts that the only solution to $a^m - b^n = 1$ in integers $a,b >0$, $m,n>1$ is $3^2 -2^3=1$, immediately giving $(x,y)=(3,2)$. Equation (2) could alternatively be approached by considering the function $f(z) = (z+1)^2 - z^x -1$ and analyzing its derivative for $z \ge 1$, showing $f$ is strictly decreasing for $x \ge 4$, thus only small $z$ are candidates. Equation (3) could be approached similarly by bounding $(z+1)^y - z^x$ and using monotonicity; the direct enumeration of small integers is simpler and avoids deeper number-theoretic machinery.