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6

I think Hardy is asking for conditions on $(a,b,c)$, to be given in terms of $(A,B,C)$ and $(\alpha,\beta,\gamma)$. The $(x,y,z)$ are dummy variables, note the "for all". For (1), which $(a,b,c)$ are orthogonal to all vectors? Answer, clearly only the zero vector, so $(a,b,c)=(0,0,0)$. For (2), which $(a,b,c)$ are orthogonal to all vectors which are ...

4

If $(x,y)$ and $(x',y')$ are two of the eleven solutions, then $x'=x+13t$, $y'=y-12t$ for some integer $t$. Hence the $11$ solutions will be of the form $x=x_0+13t$, $y=y_0-12t$ with $t$ ranging over $11$ consecutive integers, wlog over the integers $\{0,1,\ldots,10\}$. As $t=-1$ does not lead to a solution, we conclude $x_0-13<0$. As $t=11$ does not lead ...

4

Well, $M$ is the kernel of the map $$f:\mathbb{Z}^3\to\mathbb{Z}^3,\;\;\begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto\begin{pmatrix}-3x+5y+3z\\-6x+20y+9z\\-3x+25y+9z\end{pmatrix}$$ Using the standard basis $\{e_1,e_2,e_3\}$ for $\mathbb{Z}^3$, we can represent $f$ above as a matrix $$A=\begin{pmatrix}-3&5&3\\-6&20&9\\-3&25&9\end{... 4 Note c=ax+by has integer solution if and only if gcd(a,b)|c and if this exists then infinite no of integer solutions can be obtained from 1 solution by x=x_{0}+\frac{b}{d}k and y=y_{0}-\frac{a}{d}k where d is the gcd. And x_{0},y_{0} are one solution which can be obtained from euclid's algorithm. (As will jagy has mentioned) And also the ... 4 Since \gcd (152207,81103)=1111 it is the same as minimum of$$1111(137x-73y)$$Since 137x-73y=1 is solvable (say x=8 and y=15) the answer is 1111. 3 This is based on a generalization theorem and you just have your way to prove it. The theorem says: For any Diophantine equation of the form ax+by = c, it is solvable if and only if gcd(a,b) divides c. The proof will follow from your observation. 2 It means that there is no solution. Suppose on the contrary that there is a solution.$$55x+22y = 400$$then we have$$11(5x+2y)=400$$which means 11 divides 400, this is a contradiction. 2$$ \gcd( 152207, 81103 ) = ???  \frac{ 152207 }{ 81103 } = 1 + \frac{ 71104 }{ 81103 }  \frac{ 81103 }{ 71104 } = 1 + \frac{ 9999 }{ 71104 }  \frac{ 71104 }{ 9999 } = 7 + \frac{ 1111 }{ 9999 }  \frac{ 9999 }{ 1111 } = 9 + \frac{ 0 }{ 1111 } $$Simple continued fraction tableau:$$ \begin{array}{cccccccccc} & &...

2

Since I don't have the book available, nor do I recall ever reading it (at least not any recent version of it), I can't be sure what it is getting at as I don't have the context of knowing what was written in Chapter $1$. It should likely help give you an idea. However, here is what it seems to me. As it's a book about real variables, I doubt the equations ...

1

The condition for the existence of integral solutions to $ax + by = c$ is $gcd(a, b) \; | \ c$. As the set $\mathbb{N}$ is infinite so we can always find infinite numbers which aren't multiples of $gcd(a, b)$.

1

Well, you have forced the represented numbers to be odd, indeed $\pm 1 \pmod 8 \; .$ So that is it, an odd number is represented if and only if all prime factors $q \equiv \pm 3 \pmod 8$ occur with even exponents. Indeed, if there are any such $q,$ we must have $q | b$ and $q| (2c-1).$ In your example, $65 = 5 \cdot 13,$ each with odd exponent $1,$ each $5 ... 1 Hint: Plot in 2D the rectangle (real values)$0 \le x \le X$,$0 \le y \le Y$. Then plot the (real) line$x+ay=Z$. You get, in general, a triangle, a trapetium, or a rectangle minus a trapetium/triangle ($Z+1 \le x+ay$). In the second case, translate and flip the trapetium/triangle to have the right corner at the origin, and compute its (integer) ... 1 Note that$$y=\frac{1}{13}(1000-11x).$$This is saying that$y$is an integer when$1000-11x=13k$for some integer$k$, i.e., when$12\equiv11x\pmod{13}$. This happens when$x\equiv7\pmod{13}$, i.e., when$x=7+13k$for some integer$k$. Since$x$and$y$have to be positive,$k\geq0$and$$y=\frac{1}{13}(1000-11(7+13k))>0\implies k<\frac{71}{11}\implies ... 1 x+2y=1 has a solution namely x=3, y=-1. 1 Let's do a simpler example: P_n(2,3), the number of solutions to 2x+3y=n in nonnegative integers. For n>0 a solution must have either x>0 or y>0. A solution with x>0 means that (x-1,y) is a solution to 2X+3Y=n-2 so there are P_{n-2}(2,3) of these. Likewise there are P_{n-3}(2,3) solutions with y>0. But some solutions have ... 1 I would write$$5y=80-10x+1+2x$$so$$y=16-2x+\frac{1+2x}{5}$$with$$\frac{1+2x}{5}=t$$we get$$x=3t-1+\frac{1-t}{2}$\$ Can you finish?

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