0

I have come across at a proof in the book (I'm not writing it all down) but at the last lines it says: $9(73+8)^n + (73-9)8^n$ Is congruent to $9x8^n - 9x8^n$ is congruent to 0 (mod 73).

What i don't get is that how is $9(73+8)^n + (73-9)8^n$ when divided by 73 leaves the remainder $9x8^n - 9x8^n$ Or zero? I mean the value of n can be any and its binomial expansion can be bigger so how to know its really divisble by 73. Is there any simpler expression of above line so that i can understand it clearly. Or with other similar example which can be helpful to understand the concept. Thanks

Asim
  • 328
  • Because the $73$ inside the parentheses is equivalent to $0$ modulo $73$. There are $73$ different possibilities for the result of $x^n\mod 73$, after that you get $x+73\equiv x\mod 73$. – abiessu Feb 06 '21 at 20:20
  • It is not just inside parenthesis. it is about the whole expression is divisible by 73 – Asim Feb 06 '21 at 20:21
  • 1
    MathJax commands are not effective unless they are surrounded by $ signs. $9(73+8)^n$ shows up as $9(73+8)^n$ – saulspatz Feb 06 '21 at 20:22
  • 1
    Also, if you want to consider the binomial expansion, note that $(73+k)^n=73^n+73^{n-1}k\binom n1+73^{n-2}k^2\binom n2+\dots+73k^{n-1}\binom n1+k^n$ giving that every term except the last is divisible by $73$. – abiessu Feb 06 '21 at 20:25
  • Apply the Polynomial Congruence Rule in the linked dupe (or equivalently apply Sum, Product & Power Rules). It is essential to master learn these basic congruence arithmetic laws asap in order to become profiicient at modular arithmetic. – Bill Dubuque Feb 06 '21 at 20:29
  • 1
    In particular, the congruence arithmetic rules imply that given any expression composed of sums and products of integers, its value $\bmod n,$ is unchanged when we replace the arguments of the sums and products (but not any exponents) by congruent arguments. In particular this is true if we replace them by their least nonnegative reps $, a\mapsto a\bmod n$, which generally simplifies computations. – Bill Dubuque Feb 06 '21 at 20:35
  • I added another dupe link which you may find more helpful as an introduction. If you chase the links on the "divisibility by 7 rule" thread you wil find hundreds of worked examples of (all sorts) of congruence arithmetic. – Bill Dubuque Feb 06 '21 at 20:42
  • @Asim Since someone closed the question I will answer it in the comments: – Why are you reading my name Feb 06 '21 at 20:53
  • 1
    Answer: 1. $(73 + 8)^n = 73k + 8^n$ for some k since All the terms in the bionomial expansion of $(73 + 8)^n$ are divisible by $73$ except $8^n$ – Why are you reading my name Feb 06 '21 at 20:55
  • Thank u for the help . I am just getting it now. – Asim Feb 06 '21 at 20:57
  • $9(73k + 8^n) + (73 - 9)8^n = 73(9k + 8^n) + 9 \cdot 8^n - 9 \cdot 8^n$
    • – Why are you reading my name Feb 06 '21 at 20:59
  • Reduce the final part $\pmod {73}$ and you will get your answer.
    • – Why are you reading my name Feb 06 '21 at 21:00
  • . In step 2 I just opened the brackets and separated the terms so that I get all terms divisible by 73 together.
    • – Why are you reading my name Feb 06 '21 at 21:02