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Number theory is largely concerned with properties of the set of positive integers, and as such has a considerable overlap with algebra. But a simple example that illustrates the difference between a typical question in algebra and a typical question in number theory is provided by the equation $$13x − 7y = 1.$$

  • An algebraist would simply note that there is a one-parameter family of solutions: if $y = λ$ then $x = (1 + 7λ)/13,$ so the general solution is $(x,y) = ((1 + 7λ)/13, λ).$
  • A number theorist would be interested in integer solutions, and would therefore work out for which integers $λ$ the number $1 + 7λ$ is a multiple of $13.$ The answer is that $1 + 7λ$ is a multiple of $13$ if and only if $λ$ has the form $13m+ 11$ for some integer $m.$

How do we get the solution $13m + 11?$

(I have tried understanding this again and again. But where I seem to lose track is how do you jump from k= 11 to k = 13j + 11 )

Sumeet Patil
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Jul 16 '23 at 14:40
  • I guess that the above question is mostly a citation from some book. It would be a nice thing to say which one. As for the actual question (the last sentence), please reformat the question to make sure it is clear that that is what you want to know. –  Jul 16 '23 at 14:44
  • Its from princetons companion of mathematics. – Sumeet Patil Jul 16 '23 at 14:45
  • two parts: describe all $13u - 7 v = 0.$ Find two $(x,y)$ solutions to $13x - 7 y = 1.$ Take the difference of the two solutions and compare to the $(u,v)$ pairs, which make what we call a lattice. – Will Jagy Jul 16 '23 at 14:50
  • You can see why $1+7\lambda$ must be divisible by $13.$ So, we are looking for $\lambda$ such that $1+7\lambda=13k$ for some integer $k.$ Now, you know that each integer is of the form $13m+r$ for some $0≤r≤12.$ So, there are $13$ cases to consider for $\lambda.$ Considering them gives the solution $\lambda=13m+11.$ Alternatively, one can use modular arithmetic to arrive at $7\lambda \equiv_{13} 12.$ The inverse of $7$ in $\mathbb{Z}/13\mathbb{Z}$ is $2.$ This gives $\lambda\equiv_{13} 11.$ – aqualubix Jul 16 '23 at 14:56
  • $$13x-7y=1\iff13(x+1)=7(y+2)$$$$\iff x=7k-1,y=13k-2$$$$\iff x=7m+6,y=13m+11$$ for some integers $k=m+1.$ – Anne Bauval Jul 16 '23 at 15:18
  • $!\bmod 13!:\ -7y \equiv 1 \iff y \equiv \dfrac{-1}7 \equiv \dfrac{-2}{14}\equiv \dfrac{-2}1\equiv 11,$ by Gauss's algorithm. See the linked dupes for this and many other methods methods. – Bill Dubuque Jul 16 '23 at 15:41
  • 113+07=13, 013+17=7. Subtract(2) from (1) and 113-17=6. Subtract (3) from (2) and -113+27=1. x=-1 and y=-2. If you wanted positive numbers, add 713 - 137=0 giving 613 - 117 =1. Basically Euclid’s algorithm but keeping track of some equation at the same time. – gnasher729 Jul 16 '23 at 17:36
  • Note: in my prior comment, $, y\equiv 11\pmod{!13},$ means $, y = 11+ 13j,$ for some integer $,j,,$ which is precisely what you seek. – Bill Dubuque Jul 17 '23 at 03:06

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Building off of the ideas presented in @aqualubix's comment, modular arithmetic may be used to solve this Diophantine equation (without brute force). Rearranging the equation, we have $$13x=7y+1,$$ and setting $a=13x=7y+1$, we have the system of congruences $$\begin{cases} a &\equiv 0 \pmod{13},\\ a &\equiv 1 \pmod{7}. \end{cases}$$ Since $13$ and $7$ are coprime, we may use the Chinese Remainder Theorem to solve this system and ensure a unique expression for $a$ in $\mathbb{Z}/91\mathbb{Z}$. We see that $$a=7k+1$$ for some $k \in \mathbb{Z}$, and plugging this into our first congruence, we obtain $$7k+1 \equiv 0 \pmod{13} \iff k \equiv 11 \pmod{13}.$$ Hence, we have that $$k=13j+11$$ for some $j \in \mathbb{Z}$, and we obtain $$a=7(13j+11)+1 = 91j+78, \ \ j \in \mathbb{Z}.$$ However, $$91j+78=a=13x \implies x=7j+6,$$ and likewise, plugging in our formula for $a$ gives a general form for $y$ in terms of $j \in \mathbb{Z}$, which turns out to be your $$y=13j+11$$ in question.

Sean Kim
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