Suppose $n, m \in \mathbb{N}$. If $m^{6}+279=2^{n}$, find $n, m$.

What are some ways to approach this question? Instinct told me to take logarithm but doesn't work to well.

  • $\begingroup$ My first question would be: how do you know this even has solutions? $\endgroup$ – weierstrash May 30 '20 at 6:38
  • $\begingroup$ Hint: look at everything modulo 7. $\endgroup$ – achille hui May 30 '20 at 6:39
  • $\begingroup$ @achillehui From that everything becomes obvious, I'd find how to get to that ansatz a more interesting answer. $\endgroup$ – orlp May 30 '20 at 6:41
  • $\begingroup$ @orlp The general tactic is to modulo a $p$ to make those exponents as few equivalent class as possible. Since $6+1 = 7$ is a prime, $m^6$ is either $0$ or $1$. The rest is following the road.... $\endgroup$ – achille hui May 30 '20 at 7:55
  • $\begingroup$ The question looks kinda more natural with 729 instead of 279. The answer is the same, though. $\endgroup$ – Ivan Neretin May 30 '20 at 22:35

Suppose that $n$ is odd. Then

$$2=2*1^k\equiv 2*4^k=2^{2k+1}=2^n=m^6+279\equiv m^6=m^2*m^2*m^2(\text{mod }3)$$

If $m$ is divisible by $3$, then this is $0$ modulo $3$. If not, then

$$\equiv 1*1*1=1\ (\text{mod }3)$$

Since this is a contradiction either way, we conclude $n$ is even. Then we can simplify the equation to


for some $k\in\mathbb{N}$. Rearranging, we get


Now, the prime factorization of $279$ is


This means that these prime numbers ($3$ twice and $31$ once) and only these prime numbers can divide


Now, we have that



Thus, we know


However, since the other divisors are either $3$ or $9$, we may conclude


Since $k>9$ and $m>7$ imply


we only need to check a finite number of combinations for possible solutions. In fact, we find one candidate:


where $k=2$ and $m=3$. If we plug these into the other factor, we get


Since this is not divisible by $3$, we may conclude that the original equation has no solutions.

  • 1
    $\begingroup$ At the point $(2^k + m^3)(2^k - m^3) = 3\cdot 3\cdot 31$, noting that the sum of the two factors is a power of $2$ (namely $2^{k+1}$), but none of $279 + 1 = 280$, $63 + 3 = 66$, $31 + 9 = 40$ is, seems quicker/easier to me. $\endgroup$ – Daniel Fischer May 30 '20 at 12:40
  • $\begingroup$ Ahh, nice trick. That would be faster $\endgroup$ – QC_QAOA May 30 '20 at 21:07

Based on achille hui's hint we find by exhaustive computation that

$$\forall m: \left(m^6 + 279\right) \bmod 7 \in \{0, 6\}$$ $$\forall n: 2^n \bmod 7 \in \{1, 2, 4\}$$

therefore the two expressions can never be equal.

  • $\begingroup$ No need for much computation. By Fermat's Little Theorem, $m^6 \equiv (0,1) \bmod 7$ and $279=280-1$ is plainly $\equiv -1 \bmod 7$. The $1,2,4$ cycling of $2^n \bmod 7$ is observed by the time $n=4$ at most. $\endgroup$ – Keith Backman May 30 '20 at 16:43

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