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Is it true that: $13|(2^{70} + 3^{70})$ iff 13 mod(n) divides $2^{70} + 3^{70}$ mod(n)?

I assumed that it is the case. Being that correct, then 13 = 1 (mod(2)) and $2^{70} = 0$ (mod(2)); $3^{70} = 3x3x...x3 = 1x1x...x1 $ (mod(2)). Therefore $2^{70} + 3^{70}$ = 0 + 1 = 1 = 13 (mod(2)), and it follows that $13|2^{70} + 3^{70}$. I missed a lot of classes on modular arithmetic and didn't compensate it at home, so I probably got something wrong. But what I used to conceive this demonstration is that, as far as I know, the set of congruence classes forms a subring of the integers, therefore it satisfies all ring operation properties.

izzorts
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6 Answers6

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No, your approach is very false. You could take $n=1$ and the congruence would always hold.

Your statement is equivalent to $2^{70} + 3^{70}$ being congruent to $0$ modulo $13$.

Thus, what you want to do is compute $2^{70} + 3^{70}$ mod $13$. To this end you may wish to recall $a^{13}$ is $a$ modulo $13$, which is basically Fermat's Little Theorem.

quid
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  • Thanks for the answer quid. I didn't get, though, why taking $n=1$ serves as a counterexemple of the statement I assumed. Say, assume a|b, then a(mod1)=a|b(mod1)=b, surely; but if a doesn't divide b, the congruence doesn't hold. Wrong? And yes, I should've used Fermat's Little Theorem. This was a test question though, and I didn't know how to use it at the time. The argumentation I'm putting was the best I could come up with with the few things I knew then. – izzorts Nov 02 '16 at 23:22
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    It doesn't really serve as a counter example. The intent is to highlight that if what you claimed were true you could show "everything." Or, say, does $ 14$ divide $18$ because modulo $4$ we get $2$ divides $2$, which is true? Of course not. Thus, this approach is doomed. :-) – quid Nov 02 '16 at 23:36
  • Oh, now I see, thanks for this. – izzorts Nov 03 '16 at 00:00
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A trickier approach. Since $4$ and $9$ are quadratic residues $\!\!\pmod{13}$, for $p=13$ we have $$ 4^{\frac{p-1}{2}}\equiv 9^{\frac{p-1}{2}}\equiv 1\pmod{13} $$ hence $$ 4^{36}\equiv 9^{36}\equiv 1\pmod{13} $$ and $$ 2^{70}+3^{70} = 4^{35}+9^{35} \equiv 4^{-1}+9^{-1} \equiv 0\pmod{13} $$ since $\frac{1}{4}+\frac{1}{9}=\frac{\color{red}{13}}{36}$. Even simpler: $35$ is odd, hence $13=2^2+3^2$ is a divisor of $2^{2\cdot 35}+3^{2\cdot 35}$.

Jack D'Aurizio
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${\rm mod}\ 13\!:\, \left[ 2^{\large 2}\equiv -3^{\large 2}\right]^{\large 35}\!\Rightarrow\, 2^{\large 70}\equiv -3^{\large 70}\,\Rightarrow\, 2^{\large 70}+3^{\large 70}\equiv 0\ $ by the Congruence Power Rule

Community
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Bill Dubuque
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Hint. we have

$$3^3=27 \equiv 1 (13) \implies$$

$$3^{69}\equiv 1 (13) \implies$$

$$3^{70} \equiv 3 (13).$$

and

$$2^4 \equiv 3 (13) \implies$$

$$2^{12} \equiv 1 (13) \implies$$

$$2^{60} \equiv 1 (13).$$

but

$$2^{10} \equiv 10 (13)$$

then

$$2^{70} \equiv 10 (13)$$

and finally

$$2^{70}+3^{70}\equiv 10+3 (13)$$

or

$$2^{70}+3^{70} \equiv 0 (13)$$

qed.

hamam_Abdallah
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To prove this, compute $2^{70} \mod 13$ and $3^{70} \mod 13$. This can be done using Fermat's Little Theorem. Then add the results. It will be 0 mod 13 since $$ 2^{70} + 3^{70} = 13 \cdot 192550423461109399456637645953021 \, . $$

Hans Engler
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$$2^{70}=2^{12\cdot5+10}=1^5\cdot2^{10}=1024-13\cdot78\equiv 10\pmod{13}\\3^{70}=3^{12\cdot5+10}=1^5\cdot3^{10}=59049-13\cdot4542\equiv 3\pmod{13}$$ It follows $$2^{70}+3^{70}\equiv 0\pmod{13}$$

Piquito
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