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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