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By assuming that it is divisible by $65$, we get $2^{1194}+1≡0$ mod $65$. Using the fact that if $p$ and $q$ are primes $x≡y\mod p$ and $x≡y\mod q \iff x≡y\mod pq$ we get: $2^{1194}≡12 \mod 13$ and $2^{1194}≡4\mod 5$, but what's next? By Euler's Theorem I know $2^{12}≡1\mod 13$, but the exponent $1194$ is too large to compute successive squaring, how can I bring it down?

Michael Rozenberg
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Achuan Chen
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  • Hint: $1194\equiv_{12}6$. –  Jan 20 '20 at 09:54
  • You can reduce the exponent modulo $\varphi(65)=48$ – Peter Jan 20 '20 at 09:56
  • Finishing your approach, you can reduce the exponent modulo $4$ for the calculation modulo $5$ and modulo $12$ for the calculation modulo $13$ – Peter Jan 20 '20 at 10:01
  • $J$ odd, $,\color{#c00}{2^{\large 6}\equiv -1},\Rightarrow,\color{#c00}{2}^{\large \color{#c00}6J}!\equiv (\color{#c00}{-1})^{\large J}!\equiv -1,$ holds $!\bmod 5\ &\ 13,$ so $!\bmod 5\cdot 13,$ by lcm or CCRT – Bill Dubuque Jan 20 '20 at 20:46

4 Answers4

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$$2^6\equiv-1\pmod{13}$$

and $1194=1200-6=(200-1)6$

$$\implies2^{199\cdot6}=(2^6)^{199}\equiv(-1)^{199}\equiv-1$$

Similarly, $$2^{1994}=(2^2)^{997}\equiv(-1)^{997}\pmod5$$

lab bhattacharjee
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Use that $$1194=6\cdot199$$ and that $199$ is odd.

Michael Rozenberg
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$(x^{1194}+1)=(x^6+1)(x^{1188}-x^{1182}+...+x^{12}-x^6+1)$.

Now take $x=2$.

J. W. Tanner
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  • We can replace the above use of the Factor Theorem with simple modular arithmetic, viz. $${\rm if}\ , J,\ {\rm is\ odd\ then },\ !\bmod x^{\large 6}!+1!:,\ \color{#c00}{x^{\large 6}\equiv -1},\Rightarrow,\color{#c00}{x}^{\large \color{#c00}6J}!\equiv (\color{#c00}{-1})^{\large J}!\equiv -1\qquad\qquad $$ hence $\ x^{\large 6}!+1\mid x^{\large 6J}!+1\ \ \ $ – Bill Dubuque Jan 20 '20 at 21:00
  • @BillDubuque: I'm guessing your comment is what Michael Rozenberg was alluding to in his answer, though it's not spelled out there – J. W. Tanner Jan 20 '20 at 21:15
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I don't think 1194 is large at all as you can do it in 11 steps, possibly less. But I digress. One way, is to note is that by that version of Euler's theorem ( technically Fermat's theorem) $2^p\equiv 2\pmod p$ and long divide the exponent by $p=13$ giving 91 remainder 11 and use basic exponent rules and repeating to go through:

$$2^{91}2^{11}\equiv 2^72^{11}\equiv 2^6\equiv 12\pmod {13}$$

Similarly by $q=5$ :

$$2^{238}2^4\equiv 2^{47}2^32^4\equiv 2^92^22^32^4\equiv 2^12^42^22^32^4\equiv 2^12^22^3\equiv 4\pmod 5$$

now you can combine using Chinese remainder theorem.