MHB Powerful equation 5^m+7^n=k^3.

[lfdahl]

Find all the natural numbers, $m,n$ and $k$, that satisfy the equation:

\[5^m+7^n = k^3.\]

 

[Albert1]

Find all the natural numbers, $m,n$ and $k$, that satisfy the equation:

\[5^m+7^n = k^3.\]

please give a hint

 

[lfdahl]

please give a hint

Hint:

Spoiler

Step 1. $k$ must be even, so the LHS is divisible by $8$. This determines the parity of $m$ and $n$.
Step 2. Prove, that $m$ is divisible by 3. $k^3 \equiv \pm 1$ ($mod \; 7$) $\Rightarrow 5^m \equiv \pm 1$
($mod \; 7$) $\Rightarrow ...$.
Step 3. Rewrite the equation: $7^m = k^3-5^{3\cdot l} = ...$ and check for modulo $5$ and $7$ to conclude,
that ...

 

[Opalg]

Find all the natural numbers, $m,n$ and $k$, that satisfy the equation:

\[5^m+7^n = k^3.\]

[sp]Working mod $7$, $5^m \equiv k^3 \pmod7$. But a cube has to be congruent to $\pm1$ or $0\pmod7.$ A power of $5$ can never be congruent to $0\pmod7$, and $5^m \equiv \pm1\pmod7$ only if $m$ is a multiple of $3$, say $m=3r.$ The equation then becomes $k^3 - 5^{3r} = 7^n$. The left side factorises, giving $(k-5^r)(k^2 + 5^rk + 5^{2r}) = 7^n.$ So both factors on the left side must be powers of $7$, say $k-5^r = 7^s$ and $k^2 + 5^rk + 5^{2r} = 7^t$ (where $s+t=n$).

Square the first of those two equations and subtract the second equation, to get $3*5^r = 7^{2s} - 7^t$. But the left side of that equation is odd and the right side is even. So it appears that there are no solutions.

Therefore there are no natural numbers $m\ n$ and $k$ satisfying the equation $5^m+7^n = k^3.$[/sp]

 

[lfdahl]

[sp]Working mod $7$, $5^m \equiv k^3 \pmod7$. But a cube has to be congruent to $\pm1$ or $0\pmod7.$ A power of $5$ can never be congruent to $0\pmod7$, and $5^m \equiv \pm1\pmod7$ only if $m$ is a multiple of $3$, say $m=3r.$ The equation then becomes $k^3 - 5^{3r} = 7^n$. The left side factorises, giving $(k-5^r)(k^2 + 5^rk + 5^{2r}) = 7^n.$ So both factors on the left side must be powers of $7$, say $k-5^r = 7^s$ and $k^2 + 5^rk + 5^{2r} = 7^t$ (where $s+t=n$).

Square the first of those two equations and subtract the second equation, to get $3*5^r = 7^{2s} - 7^t$. But the left side of that equation is odd and the right side is even. So it appears that there are no solutions.

Therefore there are no natural numbers $m\ n$ and $k$ satisfying the equation $5^m+7^n = k^3.$[/sp]

Thankyou very much, Opalg! for a superb solution!

 

[Albert1]

[sp]Working mod $7$, $5^m \equiv k^3 \pmod7$. But a cube has to be congruent to $\pm1$ or $0\pmod7.$ A power of $5$ can never be congruent to $0\pmod7$, and $5^m \equiv \pm1\pmod7$ only if $m$ is a multiple of $3$, say $m=3r.$ The equation then becomes $k^3 - 5^{3r} = 7^n$. The left side factorises, giving $(k-5^r)(k^2 + 5^rk + 5^{2r}) = 7^n.$ So both factors on the left side must be powers of $7$, say $k-5^r = 7^s$ and $k^2 + 5^rk + 5^{2r} = 7^t$ (where $s+t=n$).

Square the first of those two equations and subtract the second equation, to get $3*5^r = 7^{2s} - 7^t$. But the left side of that equation is odd and the right side is even. So it appears that there are no solutions.

Therefore there are no natural numbers $m\ n$ and $k$ satisfying the equation $5^m+7^n = k^3.$[/sp]

Spoiler

$(k-5^r)(k^2 + 5^rk + 5^{2r}) = 7^n.$ So both factors on the left side must be powers of $7$, say $k-5^r = 7^s$ and $k^2 + 5^rk + 5^{2r} = 7^t$ (where $s+t=n$).

Square the first of those two equations and subtract the second equation, to get $3*5^r = 7^{2s} - 7^t---(A)$. But the left side of that equation is odd and the right side is even. So it appears that there are no solutions.

I think some mistakes in (A):$3*5^r = 7^{2s} - 7^t---(A)$.
k is missing (and k is even),so both sides of (A) are even

 

[Opalg]

Spoiler

$(k-5^r)(k^2 + 5^rk + 5^{2r}) = 7^n.$ So both factors on the left side must be powers of $7$, say $k-5^r = 7^s$ and $k^2 + 5^rk + 5^{2r} = 7^t$ (where $s+t=n$).

Square the first of those two equations and subtract the second equation, to get $3*5^r = 7^{2s} - 7^t---(A)$. But the left side of that equation is odd and the right side is even. So it appears that there are no solutions.

I think some mistakes in (A):$3*5^r = 7^{2s} - 7^t---(A)$.
k is missing (and k is even),so both sides of (A) are even

[sp]You are quite correct: the equation should be $3*5^rk = 7^{2s} - 7^t$. My original idea was to say that the right side of this equation is a multiple of $7$. On the left side, the only possibility then is that $k$ must be a multiple of $7$. But in that case the original equation, reduced mod $7$, says that $5^m \equiv0\pmod7$, which is impossible.

However, there is a gap in that argument, because it overlooks the possibility that $t=0$, in which case $7^{2s} - 7^t$ becomes $7^{2s} - 1$, which is not a multiple of $7$. I tried to get round this difficulty by using the "even-odd" argument in my previous comment, but that clearly does not work. All I can say in that case is that $3*5^rk = 7^{2s} - 1 = (7^s+1)(7^s-1)$, which is a multiple of $8$. But so far I have not been able to get a contradiction out of that.[/sp]

 

[lfdahl]

[sp]You are quite correct: the equation should be $3*5^rk = 7^{2s} - 7^t$. My original idea was to say that the right side of this equation is a multiple of $7$. On the left side, the only possibility then is that $k$ must be a multiple of $7$. But in that case the original equation, reduced mod $7$, says that $5^m \equiv0\pmod7$, which is impossible.

However, there is a gap in that argument, because it overlooks the possibility that $t=0$, in which case $7^{2s} - 7^t$ becomes $7^{2s} - 1$, which is not a multiple of $7$. I tried to get round this difficulty by using the "even-odd" argument in my previous comment, but that clearly does not work. All I can say in that case is that $3*5^rk = 7^{2s} - 1 = (7^s+1)(7^s-1)$, which is a multiple of $8$. But so far I have not been able to get a contradiction out of that.[/sp]

Spoiler

Thankyou very much, Albert! - for your sharp observation! I´m so sorry Opalg and Albert, that I overlooked it in Opalg´s answer. If you take a look at the suggested solution, you will find a remarkable resemblance with Opalg´s solution:

Spoiler

Let $(m,n,k)$ be a solution of the equation: $5^m+7^n = k^3$.

(1). Let us prove, that $n$ is an odd number: Indeed, $k$ must be even, and therefore the right hand side is

divisible by $8$. Since $5^m \equiv 1$ or $5$ ($mod\; 8$) for even and odd values of $m$, and

$7^n \equiv 1$ or $7$ ($mod\; 8$) for even and odd values of $n$, the only possibility is: $m$ is even, $n$ is

odd.

(2). Let us prove that $m$ is divisible by $3$: Indeed, $k$ is not divisible by $7$, otherwise $7$ divides

$5^m$. Therefore, $k^3 \equiv 1$ or $-1$ ($mod\; 7$). Thus, $5^m \equiv 1$ or $-1$ ($mod\; 7$). This is

possible only for $m = 3l$.

(3). Now we have $7^n = k^3 – 5^{3l} = (k – 5^l )(k^2 + 5^l k + 5^{2l} )$. The second factor exceeds $3$ and

therefore is divisible by $7$. If the first factor $k – 5^{m´}$ is equal to $1$, then $5^m + 7^n = k^3 \equiv 1$

($mod\; 5$) and since $n$ is odd, $7^n \equiv 1$ ($mod\; 5$), i.e. there is no solution for odd $n$.

If the first factor $k – 5^l$ is divisible by $7$, then $7$ also divides its square $k^2 − 2 · 5^{m´} k + 5^{2l}$

and since $7$ also divides the second factor $k^2 + 5^l k + 5^{2l}$ , $7$ divides their difference $3 · 5^{m´}k$.

Finally, since $5^m \not\equiv 0$ ($mod\; 7$), $7$ must divide $k$. Again, since $5^m \not\equiv 0$ ($mod\; 7$) the

equation $5^m + 7^n = k^3$ has no solution.

Thus, our equation has no solution in natural numbers (the only nonnegative integer solution is $(0,1,2)$).