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lfdahl
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Find all the natural numbers, $m,n$ and $k$, that satisfy the equation:
\[5^m+7^n = k^3.\]
\[5^m+7^n = k^3.\]
please give a hintlfdahl said:Find all the natural numbers, $m,n$ and $k$, that satisfy the equation:
\[5^m+7^n = k^3.\]
Albert said:please give a hint
[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 $s+t=n$).lfdahl said:Find all the natural numbers, $m,n$ and $k$, that satisfy the equation:
\[5^m+7^n = k^3.\]
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]
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]
[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.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
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]
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.
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.
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.
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.
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.