Kvant Math Problem 231
Consider the equation $n^x + n^y = n^z$ in natural numbers.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m59s
Source on kvant.digital
Problem
Find all solutions in natural numbers of the equation $$n^x+n^y=n^z.$$
R. Egoryan, 9th-grade student
Exploration
Consider the equation $n^x + n^y = n^z$ in natural numbers. First, observe symmetry in $x$ and $y$, allowing us to assume without loss of generality that $x \le y$. Test small values of $n$. For $n=2$, one sees that $2^1 + 2^1 = 2^2$, giving $(x, y, z) = (1,1,2)$. Trying $x=1$, $y=2$ gives $2^1 + 2^2 = 2 + 4 = 6$, not a power of 2. For $x=y$, $n^x + n^x = 2n^x = n^z$; then $2 = n^{z-x}$, so $n$ must divide 2. For $n>2$, $2$ cannot be a power of $n$, so $x=y$ leads to no solution if $n>2$. When $x < y$, factor $n^x$ to get $n^x(1 + n^{y-x}) = n^z$, so $1 + n^{y-x} = n^{z-x}$. For $n=2$, $1 + 2^k = 2^m$ is only possible for small integers; check $k=1$, $1+2=3$, not a power of 2; $k=2$, $1+4=5$, etc., suggesting $x=y$ is the only possibility for $n>1$. For $n=1$, the equation becomes $1 + 1 = 1$, impossible. The likely solution involves $x=y$ and $z=x+1$. The step most prone to error is showing no other combinations of $x<y$ yield solutions, especially when $n=2$.
Problem Understanding
We are asked to find all natural numbers $(n, x, y, z)$ satisfying $n^x + n^y = n^z$. This is a Type A problem because we must determine all solutions. The core difficulty is handling cases where $x<y$ versus $x=y$, ensuring no overlooked solutions for small $n$. The intuitive answer is that $x=y$ and $z = x+1$, giving $(n, x, y, z)$ of the form $(n, k, k, k+1)$ with $n=2$, and also $(1, x, y, z)$ is impossible. For $n>2$, only the symmetric case can work, giving $(2, k, k, k+1)$.
Proof Architecture
Lemma 1: If $n=1$, the equation has no solutions in natural numbers. True because $1^x + 1^y = 2 \ne 1^z = 1$.
Lemma 2: Without loss of generality, assume $x \le y$ by symmetry. Trivial because addition is commutative.
Lemma 3: If $x < y$, then $n^x(1 + n^{y-x}) = n^z$, so $1 + n^{y-x} = n^{z-x}$, and $n^{z-x} - n^{y-x} = 1$. This equation has no solutions for $n>1$, $y-x \ge 1$, $z-x \ge 1$. True by considering that $n^{z-x} - n^{y-x} = n^{y-x}(n^{z-y}-1) = 1$ forces $n^{y-x}=1$, impossible.
Lemma 4: If $x=y$, then $2 n^x = n^z$, implying $2 = n^{z-x}$. Only $n=2$ works, giving $z=x+1$.
Lemma 5: Check small $n$ explicitly. For $n=2$, $(x, y, z) = (k, k, k+1)$ works; for $n>2$, no solutions arise from Lemma 3.
Hardest part: proving Lemma 3 rigorously for all $n>1$ and $x<y$. The delicate point is factoring and showing no other integers solve $n^{z-x}-n^{y-x}=1$.
Solution
Assume $n$ is a natural number and $(x, y, z)$ are natural numbers satisfying $n^x + n^y = n^z$. By symmetry, take $x \le y$.
First, consider $n=1$. Then $1^x + 1^y = 1 + 1 = 2$, while $1^z = 1$, so the equation is impossible. Therefore, $n \ge 2$.
If $x = y$, then the equation becomes $n^x + n^x = 2 n^x = n^z$, so $2 = n^{z-x}$. The only solution in natural numbers occurs when $n=2$, yielding $2 = 2^{z-x}$, so $z-x=1$, giving $z=x+1$. Therefore, for $x=y$ we obtain solutions $(n, x, y, z) = (2, k, k, k+1)$ for any natural $k$.
If $x < y$, factor $n^x$ from the left-hand side:
$$n^x + n^y = n^x (1 + n^{y-x}) = n^z.$$
Divide both sides by $n^x$ to get
$$1 + n^{y-x} = n^{z-x}.$$
Let $a = y-x \ge 1$ and $b = z-x \ge 1$. Then
$$1 + n^a = n^b \implies n^b - n^a = 1 \implies n^a(n^{b-a}-1) = 1.$$
Since $n \ge 2$ and $a \ge 1$, $n^a \ge 2$, so $n^a(n^{b-a}-1) \ge 2 \cdot 1 = 2$, which cannot equal $1$. Therefore, no solutions exist with $x<y$.
All cases are exhausted. The only natural number solutions occur when $x=y$ and $n=2$, giving $z=x+1$. Therefore, the complete set of solutions is
$$(n, x, y, z) = (2, k, k, k+1), \quad k \in \mathbb{N}.$$
$\boxed{(n, x, y, z) = (2, k, k, k+1), ; k \in \mathbb{N}}$
Verification of Key Steps
In the factorization step for $x<y$, the argument $n^a(n^{b-a}-1) = 1$ relies on the minimal value $n^a \ge 2$. For $n=2$, $a=1$, $b-a \ge 1$, one gets $2(2^{b-1}-1) \ge 2$, never $1$. If a careless argument ignored the minimality, one might erroneously consider $2^1(2^{0}-1)=0$ as a solution; explicit checking prevents this.
The step $x=y$ yielding $2 n^x = n^z$ requires identifying that $2 = n^{z-x}$ forces $n=2$ since $n>2$ produces powers exceeding 2. Testing $n=3$ gives $3^{z-x} = 2$, impossible; this verifies no other $n$ works.
Explicitly testing small $k$ for $n=2$, e.g., $x=y=1$, yields $2^1 + 2^1 = 4 = 2^2$, confirming the pattern $(k,k,k+1)$ is correct.
Alternative Approaches
An alternative approach is to assume $x \le y$ and set $y = x + d$, $z = x + e$, with $d, e \ge 0$. Then $n^x(1 + n^d) = n^{x+e} \implies 1 + n^d = n^e$. From here, considering $n^e - n^d = 1$ reduces the problem to the Catalan equation $n^e - n^d = 1$, whose only solution in natural numbers is $(n,d,e) = (2,1,2)$. This immediately produces the solution $(2,k,k,k+1)$. The main approach is preferable because it does not assume knowledge of Catalan's theorem, relying only on elementary factoring and minimal value arguments, making it accessible at a 9th-grade level.