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

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In summary, the equation 5^m+7^n = k^3 can be solved by finding all the natural numbers, $m,n$ and $k$ that satisfy the equation. However, there is a gap in that argument, because it overlooks the possibility that $t=0$.
  • #1
lfdahl
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Find all the natural numbers, $m,n$ and $k$, that satisfy the equation:

\[5^m+7^n = k^3.\]
 
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  • #2
lfdahl said:
Find all the natural numbers, $m,n$ and $k$, that satisfy the equation:

\[5^m+7^n = k^3.\]
please give a hint
 
  • #3
Albert said:
please give a hint

Hint:
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 ...
 
  • #4
lfdahl said:
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]
 
  • #5
Opalg said:
[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!
 
  • #6
Opalg said:
[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]
$(k-5^r)(k^2 + 5^rk + 5^{2r}) = 7^n.$ 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
 
  • #7
Albert said:
$(k-5^r)(k^2 + 5^rk + 5^{2r}) = 7^n.$ 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]
 
  • #8
Opalg said:
[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]

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:

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 $(0,1,2)$).
 

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

What is the meaning of the equation 5^m+7^n=k^3?

The equation 5^m+7^n=k^3 is a powerful equation that represents a relationship between three numbers raised to different powers. The values of m and n determine the powers of 5 and 7 respectively, while k represents the power of the resulting sum.

What is the significance of the number 5 in the equation 5^m+7^n=k^3?

The number 5 in the equation 5^m+7^n=k^3 plays a crucial role in determining the overall value of the equation. It is the base number that is raised to the power of m, which can significantly impact the final result.

How does the value of n affect the equation 5^m+7^n=k^3?

The value of n in the equation 5^m+7^n=k^3 determines the power of 7, which when combined with the power of 5, results in the final value of k. Therefore, n plays a crucial role in determining the overall solution to the equation.

Is there a specific range of values for m and n that will make the equation 5^m+7^n=k^3 true?

Yes, there is a specific range of values for m and n that will make the equation 5^m+7^n=k^3 true. It is known as the Diophantine solution, where m and n must be positive integers that satisfy the equation. For example, m=1, n=2, and k=3 is a valid solution.

What are the real-world applications of the equation 5^m+7^n=k^3?

The equation 5^m+7^n=k^3 has various real-world applications, particularly in the field of number theory and cryptography. It is also used in the study of perfect powers and has applications in computer science and data encryption algorithms.

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