Use Fermat's Little Theorem to find the remainder of $5^{15}$ divided by $1337$.
I know Fermat's Little Theorem, but failed to understand how to use it in this.
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$$1337=7\cdot191$$ – lab bhattacharjee Dec 03 '20 at 18:05
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4Not sure Little Fermat helps much here. Just do it by Iterated Squaring. If you want, you can note that $1337=7\times 191$ and you can do each factor separately...but it is not necessary. – lulu Dec 03 '20 at 18:06
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1This thread has most of the bases covered. As lulu suggested you may want to look at this particular answer. As lab bhattacharjee suggested, using the Chinese Remainder Theorem helps also. – Jyrki Lahtonen Dec 03 '20 at 18:09
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The exponent $15$ is kinda bad for square-and-multiply, but it still beats repeated mulitplication. – Jyrki Lahtonen Dec 03 '20 at 18:10
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1@JyrkiLahtonen I think you can do $5^{16}$, then multiply by inverse of $5$. BTW thanks for all the links. – Neat Math Dec 03 '20 at 18:14
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Yeah... I don't see it. To begin with $1337=7*191$ is not prime (although we can use FLT to show $5^{15}\equiv 5^3 \equiv $ a square root of $1\pmod 7$) and $15$ is just too small in relation the $1337$ to do much good. – fleablood Dec 03 '20 at 18:22
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I wonder if they didn't mean the CRT. The CRT would save a tiny amount of effort over iterated squaring (although $5^{15}\pmod 191$ is hardly clean). And at least it can be done which $1337$ not being prime and $15 < 191$ makes using FLT rather meaningless. – fleablood Dec 03 '20 at 18:34
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$$5^3\equiv-66\pmod{191}$$ $$5^4\equiv-330\equiv52,5^5\equiv5\cdot52\equiv69,5^6\equiv5\cdot69\equiv-37,5^7\equiv5\cdot(-37)\equiv6$$
$$5^{15}\equiv5(5^7)^2\equiv5\cdot6^2\equiv-11\pmod{191}\ \ \ \ (1)$$
$$5^3\equiv-1\pmod7\implies5^{15}=(5^3)^5\equiv(-1)^5\pmod7\ \ \ \ (2)$$
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We'll forego using any 'heavy' theory and see what we can come up with to manually crank out the answer.
Anticipating what is to come we write out,
$\tag 1 \;(13 + k)10^2 \equiv k10^2 - 37 \pmod{1337}$
Getting to work,
$\tag 2 5^5 = 3125 = 3100 + 25$
Since $31 = 13\cdot2 + 5$,
$\tag 3 5^5 = (13*2 + 5) 10^2 + 25 = 2(13 + 3) 10^2 - 75$
and so by $\text{(1)}$,
$\tag 4 5^5 \equiv 2(3\cdot10^2-37)-75 \equiv 2 (263) -75\equiv 451 \pmod{1331} $
With the $\text{(4)}$ work behind us, and observing that
$\quad 451 = 450+1 \land 450 = 2 \cdot 3^2 \cdot 5^2$
we apply the binomial theorem,
$\tag 5 (2 \cdot 3^2 \cdot 5^2 + 1)^3= 2^3 \cdot 3^6 \cdot 5^6 + 3\cdot 2^2 \cdot 3^4 \cdot 5^4 + 3\cdot 2 \cdot 3^2 \cdot 5^2+ 1$
Lets us modulo reduce the first (and largest) term on the rhs of $\text{(5)}$ using mental arithmetic and $\text{(1)}$ and, at the start, $\text{(4)}$:
$\quad 2^3 \cdot 3^6 \cdot 5^6 \equiv 8 \cdot 9 \cdot 9 \cdot 9 \cdot 5 \cdot 451 =$
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot 9 \cdot 9 \cdot 2255 =$
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot 9 \cdot 9 \cdot (2300 -45) \equiv$
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot 9 \cdot 9 \cdot (1000-37 -45) \equiv$
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot 9 \cdot 9 \cdot 419 \cdot (-1) =$
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot 9 \cdot (3600 + 90 + 81) \cdot (-1) \equiv$
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot 9 \cdot (2300 -37 + 90 + 81) \cdot (-1) \equiv$
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot 9 \cdot (1000 - 37 -37 + 90 + 81) \cdot (-1) \equiv$
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot 9 \cdot 240 =$
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot (2100 + 60) \equiv$
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot (800-37 +60) \equiv $
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 9 \cdot (514) \cdot (-1) = $
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot (4500 + 90 + 36) \cdot (-1) \equiv $
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot (3200 - 37 + 90 + 36) \cdot (-1) \equiv $
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot (1900 -37 - 37 + 90 + 36) \cdot (-1) \equiv $
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot (600 - 37 -37 - 37 + 90 + 36) \cdot (-1) = $
$\quad \quad \quad \quad \quad \quad \; \; 8 \cdot 615 \cdot (-1) = $
$\quad \quad \quad \quad \quad \quad \; \; 2 \cdot 4 \cdot 615 \cdot (-1) = $
$\quad \quad \quad \quad \quad \quad \; \; 2 \cdot (2400 +60) \cdot (-1) \equiv $
$\quad \quad \quad \quad \quad \quad \; \; 2 \cdot (1160 - 37) \cdot (-1) \equiv $
$\quad \quad \quad \quad \quad \quad \; \; 2 \cdot 214 \equiv $
$\quad \quad \quad \quad \quad \quad \; \; 428 \pmod{1337}$
Next we modulo reduce the second term on the rhs of $\text{(5)}$:
$\quad 3\cdot 2^2 \cdot 3^4 \cdot 5^4 =$
$\quad \quad \quad \quad \quad \quad \; \; 3^5 \cdot 50^2 =$
$\quad \quad \quad \quad \quad \quad \; \; 3^5 \cdot 2500 \equiv$
$\quad \quad \quad \quad \quad \quad \; \; 3^5 \cdot (1200 -37) \equiv$
$\quad \quad \quad \quad \quad \quad \; \; 3^5 \cdot 174 \cdot (-1) =$
$\quad \quad \quad \quad \quad \quad \; \; 3^3 \cdot (1500 + 30 + 36) \cdot (-1)\equiv$
$\quad \quad \quad \quad \quad \quad \; \; 3^3 \cdot (200 - 37 + 30 + 36) \cdot (-1)=$
$\quad \quad \quad \quad \quad \quad \; \; 3^3 \cdot 229 \cdot (-1)=$
$\quad \quad \quad \quad \quad \quad \; \; 3^1 \cdot (1800 + 180 + 81) \cdot (-1)=$
$\quad \quad \quad \quad \quad \quad \; \; 3^1 \cdot (2000 + 61) \cdot (-1) \equiv$
$\quad \quad \quad \quad \quad \quad \; \; 3^1 \cdot 724 \cdot (-1) \equiv$
$\quad \quad \quad \quad \quad \quad \; \; 502 \pmod{1337}$
After modulo reducing the final two terms $\text{(5)}$ we can write
$\tag 6 5^{15} \equiv 451^3 \equiv 428 + 502 + 13 + 1 \equiv 944 \pmod{1337}$
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