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I am reading this note (click here and go to page 1994 for detail, in the proof of lemma 8), and found-

$$3^x + 22^y \equiv 15^z \equiv 15 \pmod{40} $$

now, I can derive $3^x + 22^y \equiv 15^z \ \pmod{40} $ but I cann't figure out how $3^x + 22^y \equiv 15^z\pmod{40}$ is $ \equiv 15 \pmod{40}$.

How to prove that the congruence has residue 15?

Consider Non-Trivial Cases
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  • This is false. With $(x,y)=(1,1)$ we have $3^1+22^1\equiv 15^2\not \equiv 15 \pmod {40}$. – lulu Aug 03 '19 at 14:30
  • @lulu Not a counterexample since $y > 1$ per the OP's prior question. – Bill Dubuque Aug 03 '19 at 14:45
  • @BillDubuque Well...but if the assumptions of the prior question apply here, then the problem is trivial, since $15^z\equiv 15\pmod {40}$ if $z$ is odd. – lulu Aug 03 '19 at 15:01
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    @lulu It is not trivial to many beginners (as witnessed by the many questions we receive on such). – Bill Dubuque Aug 03 '19 at 15:05
  • On the other hand that's a pretty advanced paper for a beginner. I'd say knowing $15^z \equiv \pm 15$ isn't some that should come to mind without thinking but that $15^z \equiv 15 \pmod{40}$ if $z$ is odd, is something any reader of this paper can verify for oneself once such a claim is made. – fleablood Aug 03 '19 at 15:55
  • FWIW $(15)^z\equiv (-1)^z\pmod 8$ and $15^z\equiv 0\pmod 3$. By Chinese remainder Th $a\equiv-1\pmod 8$ and $a\equiv 0\pmod 8$ implies $a \equiv 15\pmod {40}$ while $b\equiv 1\pmod 8$ and $a \equiv 0\pmod 8$ implies $b\equiv 25\pmod {40}$. So $15^z \equiv 15,25\pmod {40}$ depending on whether $z$ is odd or even. – fleablood Aug 03 '19 at 16:04

2 Answers2

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$ \overbrace{15(15)^{\large 2n}}^{\large 15^{\:\!\LARGE 1\:\!+\,2n}\ \ }\!\bmod 40 = \color{#0a0}5(3(\!\overbrace{15)^{\large 2n}}^{\large (-1)^{\LARGE 2n}}\!\!\bmod 8) = \color{#c00}{15}\,\ $ [so $\ 15^{\large 2k}\equiv 15(\color{#c00}{15})\equiv 25\,$ for $\,k>0\,$]

using $\ \color{#0a0}ab\bmod \color{#0a0}ac = \color{#0a0}a(b\bmod c) = $ mod Distributive Law (mDL) to factor out $\,\color{#0a0}{a = 5}$

Remark $ $ The CRT argument in fleablood's comment is essentially equivalent to the above. But - as above - the mDL is much easier to apply due to its operational form. This is explained further in the linked answer (see the "Linked" questions there for many more examples). Viewed inductively, mDL reduces the induction to the above trivial induction $\ (-1)^{\large 2n} \equiv 1\ $ (compare lulu's answer here).

Bill Dubuque
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    Note that $,z,$ is odd (say $,z = 1+2n)$ by the prior sentence in Lemma $8$ (= your prior question.) – Bill Dubuque Aug 03 '19 at 18:18
  • little confused ... $3^x + 22^y \equiv 15 \pmod{40} \implies -1 + 0 \equiv 3\pmod{4}$ is true but the paper says $3^x + 22^y \equiv 15 \pmod{40} $ has no solution when $y>1$, how is that?? – Consider Non-Trivial Cases Aug 03 '19 at 21:49
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    @Andrew reduced $!\bmod 8!:,\ 3+(-2)^{\large y}\equiv -1,\Rightarrow, (-2)^{\large y}\equiv 4,\Rightarrow, y=2,,$ contra $,y,$ odd. We don't even need the above for that since $15^{\large z}\equiv (-1)^{\large z}\equiv -1,$ by $,z,$ odd. – Bill Dubuque Aug 03 '19 at 22:27
  • ok , for $y>1$ there is no solution but from $3 + (-2)^ y \equiv -1 \pmod 8 \implies 3 -2 +1 \equiv0 \pmod 8$ which is also impossible if we put $y= 1$ then why the author said there is solution for $y=1$? – Consider Non-Trivial Cases Aug 04 '19 at 01:20
  • @Andrew They claim that there are no solutions if $y>1$ so if a solution exists (in this case) then it must have $y = 1.\ \ $ – Bill Dubuque Aug 04 '19 at 02:33
  • the author said "So $ y = 1$ when $m ≡ 3, 27 (mod 40). $" explicitly in the next line , without any "if" and further investigating $y=1$ .... should I ask this in a separate post for elaborate explanation? Please note that, I am beginner . – Consider Non-Trivial Cases Aug 04 '19 at 03:52
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    @Andrew The goal of Lemma 8 is to show that positive solutions must have $y = 1$ when $,m\not\equiv 0\pmod{!5},$ and $,m\not\equiv \ldots \pmod{120}.,$ The Lemma does not claim that a solution always exists when $,y= 1.\ \ $ – Bill Dubuque Aug 04 '19 at 04:14
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If the assumptions of your prior question hold here, then we know that $z$ is odd.

But for odd $z$, $15^z\equiv z \pmod {40}$.

To see this, We note that $15^2=225\equiv 25\pmod {40}$ and $$15^3\equiv 15\times 25\equiv 15\pmod {40}$$ Then $$15^4=15^3\times 15\equiv 15^2\equiv 25 \pmod {40}$$ and $$15^5\equiv 15^4\times 15\equiv 25\times 15\equiv 15 \pmod {40}$$ and so on.

By induction, we deduce that $15^{2n}\equiv 25\pmod {40}$ and $15^{2n+1}\equiv 15 {\pmod {40}}$.

lulu
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