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I already know some techniques to solve big exponents such as, Euler/Fermat Theorem,Euler/Carmichael,successive squaring.

But these problems seem to be more dificult as they involve operations insted of a single number.

How could I solve them ?

  • $(4^{103} + 2(5^{104}))^{102}$ by $13$
  • $53^{103}+103^{53}$ by $39$
Davood
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Goun2
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2 Answers2

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$$ (4^{103}+2 \cdot 5^{104})^{102}\overset{13}{\equiv} (4^{3 \cdot 34+1}+2 \cdot 5^{2 \cdot 52})^{102}\overset{13}{\equiv} (4^{3 \cdot 34}4^1+2 \cdot 5^{2 \cdot 52})^{102}\overset{13}{\equiv} \\ \left((4^{3})^{34} \cdot 4^1+2 \cdot (5^{2})^{52}\right)^{102}\overset{13}{\equiv} \left(64^{34} \cdot 4^1+2 \cdot 25^{52}\right)^{102}\overset{13}{\equiv} \\ \left((-1)^{34}4^1+2(-1)^{52}\right)^{102}\overset{13}{\equiv} 6^{102}\overset{13}{\equiv} 6^{6}\overset{13}{\equiv} 36^{3}\overset{13}{\equiv} 10^{3}\overset{13}{\equiv} 100.10\overset{13}{\equiv} 9.10\overset{13}{\equiv} -1 $$

$$ 53^{103}+103^{53}\overset{13}{\equiv} 1^{103}+(-1)^{53}\overset{13}{\equiv} 1-1\overset{13}{\equiv} 0; $$

$$ 53^{103}+103^{53}\overset{3}{\equiv} (-1)^{103}+1^{53}\overset{3}{\equiv} -1+1\overset{3}{\equiv} 0; $$

which implies that $53^{103}+103^{53}\overset{39}{\equiv} 0$ .

Davood
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  • why did you compute the $53^{103}+103^{53}≡0 \mod 39$? –  Aug 30 '17 at 18:49
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    @shabnamjamali ; My dear shabnamjamali, because it was asked -as the second part of question- in the first version of question, before edites makes changes. – Davood Aug 30 '17 at 19:02
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    @Famke someone asked for me to leave only one question in the comments, I'll add it again – Goun2 Aug 30 '17 at 19:04
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    @hjx , I am ok with both of them, but I think there is no need to remove one of them. – Davood Aug 30 '17 at 19:05
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    @Ok, Could you please explain a bit more the second one, The first one is a question to leave in blank in the exam, I'll focus on the second, I Have been studying this subject for only 2 weeks. – Goun2 Aug 30 '17 at 19:08
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    @hjx ; Only notice that $53\overset{13}{\equiv} 1$ and $103\overset{13}{\equiv} -1$; also notice that $53\overset{3}{\equiv} -1$ and $103\overset{3}{\equiv} 1$. Did it help you? – Davood Aug 30 '17 at 19:13
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    $λ(39) = 12$, so $53^{12} \equiv 1$ mod $39$, I got used to solving the questions this way, but $39 = 13 * 3$, I don't know what you thought to use the factorization of 39 to solve this, But as it is not an easy subject, I'll spend a time thinking to understand. – Goun2 Aug 30 '17 at 19:23
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    Anything I'll talk to you later, Thanks for the interest in helping me, I really appreciate it. – Goun2 Aug 30 '17 at 19:25
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    @hjx ; Only notice that for every prime numbers $p \neq q$ ; we have $pq \mid A$ if and only if $p \mid A$ and $q \mid A$. In other words $A\overset{pq}{\equiv}0$ if and only if $A\overset{p}{\equiv}0$ and $A\overset{q}{\equiv}0$. – Davood Aug 30 '17 at 19:35
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    @hkx Remark that the above innate $\rm\color{#c00}{sym},\color{#0a0}{metry}$ can be brought to the fore as follows

    $$\begin{align} {a,\ b} &\equiv {\color{#c00}{+c},,\color{#0a0}{-c},},\ {\rm mod},\ m,,n\ \Rightarrow\ {\rm lcm}(m,n)\mid a!+!b\[.4em] {\rm so}\ \ {53^{103},103^{53}} &\equiv {\color{#c00}{+1},,\color{#0a0}{-1},},\ {\rm mod}: 13,,3\ \Rightarrow\ {\rm lcm}(13,3)\mid 53^{103}!+!13^{53}\[.4em] {\rm because}\ {53,,103} &\equiv {\color{#c00}{+1},,\color{#0a0}{-1},},\ {\rm mod}: 13,,3\ \ {\rm and}\ \ 103,53\ \ {\rm are\ odd} \end{align}$$

     – Bill Dubuque Aug 31 '17 at 00:03
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    That explains the genesis of the divisiility. This symmetric viewpoint is explained further in this answer, including the easy proof. See also the links there for generalizations. – Bill Dubuque Aug 31 '17 at 00:03
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    @BillDubuque Thanks, let's check If I got it, 39 can be written as a product of 2 numbers, if the lcm of these 2 numbers can divide a + b, this implies that the {a,b} mod m,n are congruent to the same numbers with the difference that the signs are different. – Goun2 Aug 31 '17 at 00:49
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    @BillDubuque How did you notice that property ? by checking or by looking ? – Goun2 Aug 31 '17 at 00:52
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    @hjx It seems you have the implication reversed. See the linked answer for more details. The generalization is not difficult to see after introspecting on special cases. – Bill Dubuque Aug 31 '17 at 00:55
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    @BillDubuque I've never heard of this property do You have any book reference ? – Goun2 Aug 31 '17 at 01:08
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    @hjx I've never seen it mentioned anyhwere else. – Bill Dubuque Aug 31 '17 at 01:32
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Variants :

First congruence :

Observe first that $2$ has order $12 \bmod13$, so $4$ has order $6$, and that as $5^2\equiv -1\mod 13$, $5$ has order $4$. Thus $$4^{103}+2\cdot 5^{104}\equiv 4^{103\bmod 6}2\cdot 5^{104\bmod 4}=4^1+2\cdot 5^0=6.$$ Now $6=2\cdot 3$ and we know $2$ has order $12 \bmod 13$, whereas $3$ has order $3$. We conclude that $$6^{102}\equiv 2^{102\bmod 12}\cdot3^{102\bmod 3}=2^6\cdot3^0\equiv -1\mod 13.$$

Second congruence :

Using the Chinese remainder theorem, it is enough to calculate the expression modulo $3$ and modulo $13$.

  • Modulo $3$: $\;53\equiv 2$, which has order $2\bmod 3$, and $103\equiv 1$, so $$53^{103}+103^{53}\equiv 2^{103\bmod 2}+1^{53}\equiv 0\mod 3.$$

  • Modulo $13$: $\;53\equiv 1$ and $103\equiv -1$, so $$53^{103}+103^{53}\equiv 1+(-1)^{53}\equiv 0\mod 13.$$

The Chinese remainder theorem asserts the canonical map $$\mathbf Z/39\mathbf Z\longrightarrow\mathbf Z/3\mathbf Z\times\mathbf Z/13\mathbf Z $$ is an isomorphism, so we conclude that $\;53^{103}+103^{53}\equiv 0\mod 39$.

Bernard
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