Find the general solution $(x, y)$ ∈ $\mathbb Z × \mathbb Z$ for the system of equations $$355x − 113y = 1 \tag 1$$ $$355x + 113y = 1 \tag 2$$ You need one parameter $t ∈ \mathbb Z$ to describe the general solution.   My attempt: By Euclid's Algorithm: $\operatorname{gcd}(355,113)=1$ $1 = 355(-7)+113(22)$ So for equation $2, (x,y)=(-7,22)$. And for equation $1, (x,y)= (-7,-22)$. So the general solution is $(-7,22t)$, where $t=\{-1,1\}.$ Is my attempt correct? If not, please can you show me how do we solve this?
120906
Find the general solution $(x, y)$ ∈ $\mathbb Z × \mathbb Z$ for the system of equations $355x − 113y = 1$ ... eq$1$ $355x + 113y = 1$ ... eq$2$ You need one parameter $t ∈ \mathbb Z$ to describe the general solution.  My attempt: Euclid's Algorithm: gcd($355,113$) $= 1$ $1 = 355(-7)+113(22)$ So for eq$2$ $(x,y)=(-7,22)$ And for eq$1$ $(x,y)= (-7,-22)$ So general solution is $(-7,22t)$, where $t$={$-1,1$} Is my answer correct? If not, please how do we solve this?
90227
Solve the equation $28^x = 19^y+87^z$, where $x,y,z$ are integers.   I am confused how to find solutions for all integers and not just positive ones. How should we do that?
38113
The problem from Burton: show that the equation $n^2 + (n+1)^2 = m^3$ has no solution in the positive integers.  So far, I can see that gcd($n$,$n+1$)$=1$ and $m \equiv_4 1$ and $m=a^2 + b^2$ for some integers a,b. I'm guessing I need to reach a contradiction.  At this point, I am stuck. Any hints?
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$x^2+1=0$ in $\mathbb{Z}_7$   By trying each number, I see that there is no solution, is this correct?  And could you help me with a more direct solution, since this method is not going to work for $\mathbb{Z}_p$ when $p$ is a large prime number .  Thank you very much!
13074
I have to see for which value of $a$ this system is solvable: $$ \begin{cases}   3x \equiv a\pmod{28}\\   ax \equiv4\pmod{21}\\ \end{cases} $$ I check the two congruences separately: $$ \gcd(3,28)=1\mid a\;\checkmark $$ $$ \gcd(a,21)\mid4 \iff \gcd(a, 21) = 1 \iff a \not\equiv 0\;(7)\;\land a \not\equiv 0\;(3) $$  $$ \begin{cases}   3x \equiv a\pmod{4}\\   ax \equiv4\pmod{3}\\ \end{cases}\\ \gcd(4,3) = 1\;\checkmark $$ $$ \begin{cases}   3x \equiv a\pmod{7}\\   ax \equiv4\pmod{7}\\ \end{cases} $$ and from here I can't go on.  So, my problem is that I can't figure out how to check if a system of congruences is solvable.  EDIT:  maybe I found the solution: $ \gcd(7,7) = 7\mid a-4 \iff a = 11+7k \land a \not\equiv 0\;(7)$
93879
The problem I am given is finding a solution to the following pair of equations:  $3x + 4y$ $\equiv$ $5$ $mod$ $13$  $2x + 5y$ $\equiv$ $7$ $mod$ $13$  By reading the methodology here:   Since the modulos in my problem are also the same, I am able to use substitution, but this leaves fractions in my expressions.  I then used matrix multiplication as follows:  $\begin{bmatrix} 3 &amp; 4\\ 2 &amp; 5\end{bmatrix}$ $\begin{bmatrix}\ x\\ y\end{bmatrix}$ = $\begin{bmatrix}\ 5\\ 7\end{bmatrix}$   And solving this also gives me a matrix with fractions, so I am a bit lost on how to solve this problem.  Any help is appreciated. Thank you very much.
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I know it might be too easy for you guys here. I'm practicing some problems in the textbook, but this one really drove me crazy. From $\gcd( a, b ) = 1$, I have $ax + by = 1$, where should I go from here? The extra $ac$ is so annoying. Any hint?    Thanks, Chan
58186
Help me understand problems of this type a bit more intuitively.   The solution $C(3+11−1,11)$ seems simple enough, but I got stuck thinking about how many integers you are choosing from within $x_1$, $x_2$, and $x_3$ and I was trying to get an $n$ - as in $C(n+r-1,r)$ - this way. Also, I though the $r$ was 3 since you have $x_1$, $x_2$, $x_3$ choices. Basically, I went wrong at every step. Help!
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If $x,y\in\mathbb Z$, solve $$x^2+xy+y^2=x^2y^2$$  We could try some factorizations. $x^2+y^2=xy(xy-1)$. We may as well add $2xy$ to both sides: $(x+y)^2=xy(xy+1)$. Then we could subtract $x^2y^2$ from both sides: $(x+y+xy)(x+y-xy)=xy$  Or we could somehow use the fact that if an integer is between two consecutive integer squares, then it can't be a square itself.  I'd like to get some more help here, some other strategies if these can't be used. Thanks.
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My question is this:      $y^2 = \frac{x^5 - 1}{x-1}$ &amp; $x,y \in \mathbb{Z}$     Source: BdMO 2016 Regionals Set 1 Question 3.   I think this is the exact same question. Actually I didn't get the solution in the link. Please don't mark as duplicate.  I tried solving this same problem by writing code; there seems to be $6$ solution pairs to this equation. How do I get the $6$ solutions mathematically? I indeed need a full solution. Please help.
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I know how to solve $ax + by = c$ using Extended Algorithm. But with more than variables, I'm lost :(.   To verify if it has an integer solution is easy, since we only need to check for $\gcd(a,b,c)|d$. Other than that, how can we find an integer solution for this equation?    Thanks, Chan
77243
I'm trying to learn how to use the Chinese Remainder Theorem (CRT), and in order to give some context:  We search for all $x ∈ Z$, where $Z$ is the set of integers.  $x≡a_1\pmod{m_1}$  $x≡a_2\pmod{m_2}$  ...  $x≡a_k\pmod{m_k}$  The easy case (which I can solve) is if all $m_i$, where $i=1,2,...,k$ are pairwise coprime.  Example:  $x≡4\pmod 5$  $x≡5\pmod 6 $  $x≡3\pmod 7$  Then the first equation is satisfied iff $x=4+5s$, for some $s ∈ Z$.  These $x$ also satisfy the second equation iff $4+5s≡_6 5 ↔ -s≡_6 1 ↔ s=-1+6t$, for some $t ∈ Z$. Thus $x=4+5(-1+6t)=-1+30t$.  Lastly, these $x$ also satisfy the third equation iff $-1+30t ≡_7 3 ↔ 2t ≡_7 4 ↔ t ≡_7 2 ↔ t = 2+7n$, for some $n ∈ Z$. Thus $x=59+210n$.  Now to my issue, I have the problem:  $x≡2\pmod 4$  $x≡3\pmod 5$  $x≡5\pmod 6$  Here $\gcd(4,6)=2$, so they are not coprime and I don't know how to solve this. Can someone please solve it and explain why the problem becomes more difficult to solve when $m_i$ are not pairwise coprime.
68409
Solve the system of congruence relations:  $2x+3y\equiv 1\pmod {11}$  $x+4y\equiv 4\pmod {11}$  Could someone give a hint how to solve this system.  I know that Chinese remainder theorem can't be used because modulo numbers ($11$) are not relatively prime.
14282
I was presented with the following problem:      Prove that there exist solutions to $x^2+y^2=z^n$ for all $n$, with $x,y,z, n \in \mathbb{N}$   I showed that by taking any Pythagorean triple $x^2+y^2=z^2$ and multiplying by $z^{2(n-1)}$ we get $(z^{n-1}x)^2+(z^{n-1}y)^2=(z^2)^n$, which allows me to generate solutions easily for any value of $n$. I managed to find several similar questions on this site, such as  concerning the specific case $n=3$. I notice that all of these questions take a similar approach and start with a Pythagorean triple and use it to generate general solutions.  Is there a way to prove the statement (or better yet provide solutions to the equation) without first relying on Pythagoras?
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I'm use to finding the solutions of linear Diophantine equations, but what are you suppose to do when you have quadratic terms? For example consider the following problem:  Find all solutions in positive integers to the following Diophantine equation $x^2 + 2y^2 = z^2$  I'd usually start by finding the gcd and use some other tricks, but I'm not sure how to approach this type of problem
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Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $B ∈ \mathbb{Q}^m$. Suppose that the system of linear equations $AX = B$ has a solution in $\mathbb{R}^n$. Does it necessarily have a solution in $\mathbb{Q}^n$?  Where do I start?    I feel I can use this to :  $$x_i=\frac{1}{a'_{ii}}\left(b'_i - \sum_{j=i+1}^{k} a'_{ij} x_j \right).$$    Here is a thought that just crossed my mind, but I'm not sure if this is legal:  Suppose $Y\in \{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $Y'\in\mathbb{Q}^n$ are solutions to $AX=B$. Then $AY=AY'=B$, or $A(Y-Y')=A(Y+(-Y'))=0$, but  addition between $\{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $\mathbb{Q}^n$ is . . .
88047
Find one integer solution to the Diophantine equation   \begin{equation*} 18x+14y+63z=5. \end{equation*}   If this were only a linear equation over $\mathbb{Z}^2$, then I could easily solve it by using the extended Euclidean algorithm... but I have no idea how to do this with more than 2 unknowns...