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Let $N = 3^{105} + 4^{105}$. Show that $7 \vert N.$

So the rule for a number to be divisible by $7$ is that we have to take the last digit and double it and see if the difference between that and the original number is divisible by $7$.

This doesn't seem to be the way to go here since the number is huge. Using computer I managed to compute $N \equiv 7 \pmod{10}$, thus $7$ is the last digit of $N$. Now I would need to determine if $3^{105} + 4^{105} - 14$ is divisible by $7$ which isn't that easy to do. Also, I wasn't able even to compute the last digit by hand here, there seems to be an alternative way to go about this?

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    "I managed to compute $n\equiv 7\pmod{10}$" In case it wasn't clear, the last digit is completely irrelevant to this problem. – JMoravitz Oct 19 '20 at 12:41
  • It is indeed. However, the last digit is something usually associated when being asked about divisibility by $7$. –  Oct 19 '20 at 12:42
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    Fair enough I suppose... if you are referring to the divisibility rule for $7$ that $\overline{a_1a_2a_3\dots a_n} \equiv \overline{a_1a_2\dots a_{n-1}} - 2a_n\pmod{7}$, equivalently $\dfrac{\overline{a_1a_2\dots a_{n-1}a_n} - a_n}{10}-2a_n\pmod{7}$ but then you should have been looking at $\dfrac{3^{105}+4^{105}-7}{10}-14$ being divisible by $7$ which looks worse than the original problem. Note, the division by ten and subtraction by $7$ that you missed. – JMoravitz Oct 19 '20 at 12:46
  • $3^{105}+4^{105}=(3+4)(3^{104}-3^{103}4+\cdots-3\cdot4^{103}+4^{104})$ – J. W. Tanner Oct 19 '20 at 13:45

1 Answers1

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$$4 \equiv -3 \pmod 7$$

$$N = 3^{105} + 4^{105} \equiv 3^{105} + (-3)^{105} \equiv 0 \pmod 7$$

cosmo5
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