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This question already has an answer here:

Let $A=3^{105} + 4^{105}$. Show that $7\mid A$.

My attempt-

I tried to convert $3^{105}$ like this: $3^{105}\equiv 3\cdot(3^2)^{52}\equiv 3\cdot 9^{52}\equiv 3\cdot2^{52} \pmod 7$. And for $4^{105}$ I tried the same way to convert it into $4\cdot 2^{52}$ but that's not possible and I'm getting $4^{79}\cdot 2^{52} \pmod 7$. I had the idea of doing it in $3^{105} + 4^{105}\equiv 3\cdot2^{52}+4\cdot 2^{52}\equiv 7\cdot2^{52}\equiv 0 \pmod 7$.

But I'm getting $4^{79}$ making it complex.

Can anyone suggest me some easier method? Please.

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marked as duplicate by Bill Dubuque elementary-number-theory Jun 16 '17 at 19:37

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    $\begingroup$ If $n$ is odd, $x^n + y^n$ has a factor of $x + y$. It's an identity. $\endgroup$ – quasi May 2 '17 at 5:19
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    $\begingroup$ For example, you know how to factor $x^3+y^3$, right? How about $x^5 + y^5$? $\endgroup$ – quasi May 2 '17 at 5:21
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    $\begingroup$ Isn't $4\equiv-3\pmod7$? $\endgroup$ – Lord Shark the Unknown May 2 '17 at 5:22
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    $\begingroup$ Because you want to prove that $4^{105}\equiv-3^{105}\pmod7$. $\endgroup$ – Lord Shark the Unknown May 2 '17 at 5:33
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    $\begingroup$ Taking quasi's hint one farther: $x^{105}+y^{105}=(x+y)(x^{104}-x^{103}\cdot y + \cdots - x \cdot y^{103}+y^{104})\,$. $\endgroup$ – dxiv May 4 '17 at 1:03
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You can convert $4^{105}$ into $4\cdot2^{52}$ mod $7$, since $4^2=16\equiv2$ mod $7$:

$$4^{105}=4\cdot4^{2\cdot52}=4\cdot16^{52}\equiv4\cdot2^{52}$$

This gives you

$$3^{105}+4^{105}\equiv3\cdot2^{52}+4\cdot2^{52}=7\cdot2^{52}\equiv0\mod 7$$

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The integer $A=3^{105}+4^{105}=3^{105}-(-4)^{105}$ is a multiple of $7=3-(-4)$, because $x-a$ is always a factor of the polynomial $x^{n}-a^{n}$, $n\in \mathbb{N}$.

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By the binomial theorem, $4^{105}=(7-3)^{105}=7a-3^{105}$.

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${\rm mod}\ 7\!:\,\ 4\equiv -3\ \Rightarrow\ \color{#c00}{4^{\large 105}}\equiv(-3)^{\large 105}\equiv\color{#0a0}{-3^{\large 105}}\ $ by the Congruence Power Rule.

Therefore $\ A = 3^{\large 105} + \color{#c00}{4^{\large 105}}\equiv\, 3^{\large 105} \color{#0a0}{- 3^{\large 105}}\equiv\, 0\ $ by the Congruence Sum Rule.

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