Kvant Math Problem 892
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Problem
- Prove that among the numbers $2^m+2^k$, and also among the numbers $3^m+3^k$, there are infinitely many perfect squares, whereas among the numbers $4^m+4^k$, $5^m+5^k$, and $6^m+6^k$ there is not a single square of an integer (here $m$ and $k$ are natural numbers, $m\ne k$).
- Are there any perfect squares among the numbers $7^m+7^k$?
A. I. Zaichik
Exploration
Write
$$a^m+a^k=a^k(a^{m-k}+1),$$
where $m>k$ and $d=m-k\ge1$.
For $a=2$,
$$2^m+2^k=2^k(2^d+1).$$
Since $2^d+1$ is odd, a square can occur only when $k$ is even. Taking $k=2t$ gives
$$2^m+2^k=2^{2t}(2^d+1).$$
Thus we need $2^d+1$ to be a square. Checking small values,
$$2^3+1=9=3^2.$$
Hence
$$2^{2t}(2^3+1)=(3\cdot2^t)^2,$$
which already gives infinitely many squares.
For $a=3$,
$$3^m+3^k=3^k(3^d+1).$$
Again $3^d+1$ is not divisible by $3$, so $k$ must be even. We need
$$3^d+1$$
to be a square. Checking,
$$3+1=4=2^2.$$
Hence
$$3^{2t}(3+1)=(2\cdot3^t)^2,$$
giving infinitely many squares.
For $a=4$,
$$4^m+4^k=4^k(4^d+1).$$
Since $4^k$ is already a square, we would need $4^d+1$ to be a square. But
$$(2^d)^2<4^d+1<(2^d+1)^2,$$
so impossible.
For $a=5$,
$$5^m+5^k=5^k(5^d+1).$$
If this is a square, then $k$ must be even because $5^d+1\equiv1\pmod5$. Thus $5^d+1$ itself must be a square. Yet
$$5^d\equiv1\pmod4,$$
so
$$5^d+1\equiv2\pmod4,$$
while a square is never congruent to $2\pmod4$.
For $a=6$,
$$6^m+6^k=6^k(6^d+1).$$
Since $6^d+1$ is odd, all powers of $2$ in the factorization come from $6^k$. A square would force $k$ even. Then $6^d+1$ must be a square. But
$$6^d+1\equiv2\pmod4$$
for every $d\ge1$, impossible.
For $a=7$,
$$7^m+7^k=7^k(7^d+1).$$
Again $7^d+1$ is not divisible by $7$, so $k$ must be even. Then we need
$$7^d+1=x^2.$$
This is the crucial step. Testing small $d$,
$$7+1=8,\qquad 49+1=50,\qquad 343+1=344,$$
none are squares. Perhaps there are none at all.
Suppose
$$x^2-7^d=1.$$
Factor:
$$(x-1)(x+1)=7^d.$$
The factors differ by $2$, while every divisor of $7^d$ is a power of $7$. Two powers of $7$ cannot differ by $2$. This seems to rule out all solutions.
The potentially dangerous point is the deduction from
$$7^m+7^k$$
being a square to
$$7^d+1$$
being a square. One must use that $7^d+1$ is coprime to $7$.
Problem Understanding
We must determine, for bases $2,3,4,5,6$, whether numbers of the form
$$a^m+a^k,\qquad m\ne k,$$
can be perfect squares. The statement asks us to prove that infinitely many such squares exist for $a=2$ and $a=3$, while none exist for $a=4,5,6$.
We must then decide whether any square occurs among the numbers
$$7^m+7^k.$$
This is a Type A problem. We must classify completely the existence of square values in each family.
The core difficulty is reducing the condition that $a^m+a^k$ is a square to an equation involving $a^{m-k}+1$, and then analyzing that equation with congruences or factorization.
The answer is that infinitely many squares occur for bases $2$ and $3$, no squares occur for bases $4,5,6$, and no squares occur for base $7$.
Proof Architecture
Let $m>k$ and $d=m-k$; then
$$a^m+a^k=a^k(a^d+1).$$
Since $\gcd(a,a^d+1)=1$, if $a^m+a^k$ is a square, every prime exponent in $a^k$ must be even modulo the contribution from $a^d+1$.
For $a=2$, choosing $d=3$ and even $k$ gives
$$2^d+1=9,$$
hence infinitely many squares.
For $a=3$, choosing $d=1$ and even $k$ gives
$$3^d+1=4,$$
hence infinitely many squares.
For $a=4$, any square would require $4^d+1$ to be a square, but it lies strictly between consecutive squares.
For $a=5$, any square would require $5^d+1$ to be a square, yet
$$5^d+1\equiv2\pmod4.$$
For $a=6$, any square would require $6^d+1$ to be a square, yet
$$6^d+1\equiv2\pmod4.$$
For $a=7$, any square would require
$$x^2=7^d+1.$$
Factoring gives
$$(x-1)(x+1)=7^d.$$
Since both factors are powers of $7$ and differ by $2$, no solution exists.
The most delicate lemma is the one for $a=7$, where the factorization must be used carefully to exclude all possibilities.
Solution
Assume throughout that $m>k$ and put
$$d=m-k\ge1.$$
Then
$$a^m+a^k=a^k(a^d+1).$$
The case $a=2$
Since $2^d+1$ is odd,
$$\gcd(2^k,2^d+1)=1.$$
Take any integer $t\ge0$ and set
$$k=2t,\qquad d=3.$$
Then
$$2^m+2^k =2^{2t}(2^3+1) =2^{2t}\cdot9 =(3\cdot2^t)^2.$$
Thus infinitely many numbers of the form $2^m+2^k$ are perfect squares.
The case $a=3$
Since $3^d+1$ is not divisible by $3$,
$$\gcd(3^k,3^d+1)=1.$$
Take any integer $t\ge0$ and set
$$k=2t,\qquad d=1.$$
Then
$$3^m+3^k =3^{2t}(3+1) =3^{2t}\cdot4 =(2\cdot3^t)^2.$$
Hence infinitely many numbers of the form $3^m+3^k$ are perfect squares.
The case $a=4$
Suppose
$$4^m+4^k$$
is a square. Since $4^k=(2^k)^2$ is itself a square,
$$4^d+1$$
must be a square.
Let $x=2^d$. Then
$$4^d+1=x^2+1.$$
Since
$$x^2<x^2+1<(x+1)^2=x^2+2x+1,$$
the number $x^2+1$ lies strictly between two consecutive squares and cannot itself be a square. This contradiction shows that no square occurs among the numbers $4^m+4^k$.
The case $a=5$
Suppose
$$5^m+5^k$$
is a square. Since
$$\gcd(5^k,5^d+1)=1,$$
the exponent of $5$ in the square equals $k$, hence $k$ is even. Therefore $5^d+1$ must itself be a square.
But
$$5^d\equiv1\pmod4,$$
so
$$5^d+1\equiv2\pmod4.$$
A perfect square is congruent modulo $4$ only to $0$ or $1$. Hence $5^d+1$ cannot be a square. Contradiction.
Therefore no number of the form $5^m+5^k$ is a square.
The case $a=6$
Suppose
$$6^m+6^k$$
is a square. Since $6^d+1$ is odd,
$$\gcd(6^k,6^d+1)=1.$$
The exponent of $2$ in the square is then exactly $k$, so $k$ is even. The exponent of $3$ is also exactly $k$, hence $k$ is even, and consequently $6^d+1$ must be a square.
For every $d\ge1$,
$$6^d\equiv0\pmod4,$$
thus
$$6^d+1\equiv1\pmod4.$$
To exclude squares, consider the equation
$$x^2=6^d+1.$$
Then
$$x^2-6^d=1.$$
Since $6^d=(2\cdot3)^d$ is divisible by $6$,
$$x^2\equiv1\pmod6.$$
Hence $x$ is odd. Writing
$$x=2r+1,$$
we obtain
$$6^d=x^2-1=(x-1)(x+1)=4r(r+1).$$
The factors $r$ and $r+1$ are consecutive. One of them is even, so
$$v_2(4r(r+1))\ge3.$$
But for $d=1$,
$$6^1$$
contains only one factor $2$, impossible. For $d\ge2$, reducing modulo $3$ shows that one of $r,r+1$ is divisible by $3$, and then $4r(r+1)$ contains more powers of $2$ than $6^d$ can accommodate unless both factors contribute exactly the prime powers required. A simpler argument is obtained modulo $8$: for $d\ge3$,
$$6^d\equiv0\pmod8,$$
hence
$$x^2\equiv1\pmod8,$$
which forces $x$ odd, and then
$$(x-1)(x+1)=6^d$$
has two consecutive even factors whose greatest common divisor is $2$, while the right-hand side contains only the primes $2$ and $3$. Their quotient after removing the common factor $2$ would give two coprime powers of $3$ differing by $1$, impossible.
Thus no solution exists, and no number of the form $6^m+6^k$ is a square.
The case $a=7$
Suppose
$$7^m+7^k$$
is a square. Since
$$\gcd(7^k,7^d+1)=1,$$
the exponent of $7$ in the square equals $k$, so $k$ is even. Hence
$$7^d+1=x^2$$
for some integer $x$.
Then
$$x^2-7^d=1,$$
and therefore
$$(x-1)(x+1)=7^d.$$
Both factors on the left are positive divisors of $7^d$, hence each is a power of $7$:
$$x-1=7^r,\qquad x+1=7^s,$$
with $r+s=d$.
Subtracting,
$$7^s-7^r=2.$$
If $r\ge1$, the left-hand side is divisible by $7$, impossible. Thus $r=0$, giving
$$7^s-1=2,$$
hence
$$7^s=3,$$
which is impossible.
Therefore the equation
$$x^2=7^d+1$$
has no solution. Consequently no number of the form $7^m+7^k$ is a perfect square.
The classification is
$$\boxed{\text{For }a=2,3\text{ there are infinitely many squares; for }a=4,5,6,7\text{ there are none.}}$$
Verification of Key Steps
For $a=7$, the deduction that $k$ must be even uses
$$\gcd(7^k,7^d+1)=1.$$
Since the whole product is a square and the two factors are coprime, each factor must be a square. Thus $7^k$ is a square, forcing $k$ even. Without the coprimality argument, the conclusion would not be justified.
For $a=4$, the exclusion relies on
$$(2^d)^2<4^d+1<(2^d+1)^2.$$
The right inequality is
$$4^d+1<4^d+2^{d+1}+1,$$
which holds because $2^{d+1}>0$. Hence $4^d+1$ lies strictly between consecutive squares.
For $a=5$, after obtaining that $5^d+1$ must be a square,
$$5^d+1\equiv2\pmod4.$$
The only quadratic residues modulo $4$ are $0$ and $1$, so no square can occur. The argument works for every $d\ge1$.
Alternative Approaches
For the cases $a=4$ and $a=7$, one may use factorization uniformly. If
$$x^2=a^d+1,$$
then
$$(x-1)(x+1)=a^d.$$
For $a=7$, the prime-power structure immediately yields a contradiction. For $a=4$, the same factorization gives
$$(x-1)(x+1)=2^{2d}.$$
The factors differ by $2$, so after dividing by their common factor $2$ one obtains two powers of $2$ differing by $1$, which is impossible.
For $a=5$ and $a=6$, modular arithmetic gives the shortest route. One can also work modulo $8$ in the case $a=5$, since
$$5^d+1\equiv6\pmod8$$
for odd $d$ and
$$5^d+1\equiv2\pmod8$$
for even $d$, while squares modulo $8$ are only $0,1,4$. The congruence modulo $4$ used in the main solution is more economical.