New answers tagged diophantine-equations
0
If $p$ is an odd prime and $-34\equiv a^2\pmod{p}$ for some $a$ then
$$0^4-17\equiv2\left(\tfrac{a}{2}\right)^2\pmod{p}.$$
The linked question gives solutions to the congruence
$$a^4-17b^4\equiv2c^2\pmod{p},$$
for every prime number $p$. Moreover, all solutions given there have $b\not\equiv1\pmod{p}$. So for
$$x\equiv ab^{-1}\pmod{p}
\qquad\text{ and }\...
5
Below is a proof that the following is the full list of solutions:
$$\begin{array}{rrr}
p&m&n\\
\hline
2&3&1\\
2&16&7\\
3&9&2\\
3&22&6\\
5&10&1\\
7&147&19\\
11&121&9\\
\end{array}$$
Let $m$ and $n$ be positive integers and $p$ a prime number such that
$$\frac{m(m+5)}{n(n+5)}=p^2.$$
Setting $a:=...
2
Numerically:
? for(m=2,10^5,for(n=1,m-1,p2=m*(m+5)/(n*(n+5));if(issquare(p2)&&p2==floor(p2),p=sqrtint(p2);if(isprime(p),print1("("n","m","p"), ")))))
(1,3,2), (2,9,3), (1,10,5), (7,16,2), (6,22,3), (9,121,11), (19,147,7),
Let $n$ is parameter, $2m+5\to x$, $2p\to y$, then solving of Pell equation $x^2-n(n+5)y^2=25$ can help us filter out solutions ...
2
The best source on this is HURWITZ 1907. A proof for the Markov numbers is included. Lots of students on this site are aware of a contest technique called "Vieta Jumping." The improvement that Hurwitz gives is to emphasize the geometry of "fundamental solutions" of one of his diophantine equations, where we just need to show existence of solutions obeying ...
2
This does not answers your question in full, it is just a start of the analysis of the problem and it can be a way to start.
From $\dfrac{m(m+5)}{n(n+5)}=p^2$ we obtain $m(m+5)=p^2n(n+5)$.
If you view this as a quadratic equation in $n$ we obtain:
$p^2n^2+5p^2n-m(m+5)=0$.
The solutions are $n=\dfrac{-5p^2+\sqrt{25p^4+4p^2m(m+5)}}{2p^2}$
The expression ...
0
Easy to solve in $$2x+5y=7z$$ as, $$x=y=z$$ is a solution trivially. Adding 5 to x, and subtracting 2 from y, also adds 0. Adding or subtracting 0 is idempotent. Adding or subtracting any solutions ( for x and y at least) gives a new solution, for this same reason (we are adding or subtracting 0 mod 7).We Also have adding 6 to x and subtracting 1 from y ...
9
If it is reducible over $\Bbb{Z}$ then it has a root in $\Bbb{Z}$, say $k\in\Bbb{Z}$. Then $k^3+nk+1=0$ so
$$-1=k^3+nk=k(k^2+n),$$
which shows that $k$ divides $-1$, so $k=\pm1$. Solving the two equations
$$1^3+n\cdot1+1=0\qquad\text{ and }\qquad (-1)^3+n\cdot(-1)+1=0,$$
yields $n=-2$ and $n=0$ as the only values for which the polynomial is reducible over $\...
0
Since $A$ may be any odd number $\ge3$ and we can find one or more triples for any odd leg $A\ge 3$ using a function of $(m,A).$
We know that $A=m^2-n^2\implies n=\sqrt{m^2-A}.$
$$\text{We can let }n=\sqrt{m^2-A}\text{ where }\lceil\sqrt{A}\space\rceil\le m\le \frac{A+1}{2}$$
$\text{Note: }n\in \mathbb{R}\implies \sqrt{A}\lt m\land n<m\implies m\le\frac{...
3
Hint
Since it is
$$
3^{\,n} x - 2^{\,n} y = 3^{\,n} x + 2^{\,n} \left( { - y} \right) = 1 = \gcd \left( {2^{\,n} ,3^{\,n} } \right)
$$
then the Bezout Identity assures that we can find two integers such that the identity above is satisfied.
The two integers $x,y,$ will come by applying the Extended Euclidean Algorithm to the couple $3^n , \, 2^n$, or by ...
2
Lets see the result needs to be odd ( 1 mod 2) . That requires x be 1 mod 2 . The result is also, 1 mod 3 leading to if n is odd y is 2 mod 3, if n is even y is 1 mod 3 . These two combine because the result is Also 1 mod 6 (1 mod 2 and 1 mod 3 together) etc. It's Also 1 mod n but unless n is prime we don't get much. if we say n is prime, we get x and y ...
0
Bézout's theorem states that if $\gcd(a,b)=d$, then exists some $p,q \in \mathbb{Z}$ such that:
$$ap+bq=d$$
then, you have to multiply by $\frac{n}{d}$:
$$a\left( \frac{pn}{d} \right) + b\left( \frac{qn}{d} \right) = n$$
and notice that:
$$a\left( \frac{pn}{d}{{\color{red} {-1}}} \right) + b\left( \frac{qn}{d}+{{\color{red} {\frac{a}{b}} }} \right) = n$...
5
This is a special case of Mihăilescu's theorem (a.k.a. Catalan's conjecture) that has been illustrated here many times before. There are no nontrivial solutions to
$$2^x-3^y=1,$$
nontrivial meaning $x,y\geq2$.
Proof: Clearly $x,y\geq0$. If $x$ is even, say $x=2z$, then
$$3^y=2^x-1=(2^z-1)(2^z+1),$$
and so the two factors on the right hand side are two ...
2
If you look at how to get from one of your solutions to the next, note that $x$ increases by $6$ and $y$ decreases by $5$. In fact, for any $n$, with one solution $(x,y)$, all other solutions are given by $(x+6k, y-5k)$ for some integer $k$.
If all four solutions should have $x,y\geq1$, then because we want the minimal possible case, one solution ought to ...
1
Constructive solution: given a fraction $\frac xy$, choose a non-zero number $k$. Then define $z=-k^2x$ and $t=ky$, thus
$$\frac zt =\frac{-k^2x}{ky}.$$
Then either way their sum will be $(1-k)\frac xy$.
An example with small numbers would be $x=1$, $y=2$ and $k=2$ which gives
$$\frac12+\frac{-4}{4}=-\frac12.$$
NB: You may even construct more "exotic" ...
1
This can be written as,
$$\frac{xt+yz}{yt} = \frac{x+z}{y+t} \implies xty + y^2z + yzt +xt^2 = ytx + yzt$$
$$y^2z+xt^2 = 0$$
1
I got the equation $$y^2z+xt^2=0$$
0
Above equation shown below,
$x^2-xy+y^2=13$ -----$(1)$
Equation $(1)$ has parametric solution:
$x=w(3k^2+2k-4)$
$y=w(4k^2-6k-1)$
Where, $w=[(1)/(k^2-k+1)]$
For $k=2$, we get $(x,y)=(4,1)$
4
If both $x,y$ are positive you can use the inequality $x^2 + y^2 \ge 2xy$. We have:
$$13 = x^2 + y^2 - xy \ge 2xy - xy = xy$$
Thus we have $xy \le 13$. Now WLOG let $x \ge y$. Then we have that $y \le \sqrt{13}$. Thus $y=1,2$ or $3$.
Plug these into the equation to check for solutions. Also don't forget that if $(x,y)$ is a solution, so is $(y,x)$.
Note ...
4
Hint: $x^2-xy+y^2=\frac34(x-y)^2+\frac14(x+y)^2$, so you want to solve $3(x-y)^2+(x+y)^2=52$ for integers $x,y$. Start by bounding $\lvert x-y\rvert$ and $\lvert x+y\rvert$.
1
I would write $$x_{1,2}=\frac{y}{2}\pm \sqrt{13-\frac{3}{4}y^2}$$
We consider here the equation $$x^2-xy+y^2-13=0$$ as an equation in $x$ using the quadratic formula. So we get $$|y|\le \sqrt{\frac{52}{3}}$$ this means $$|y|\le 4$$
It must be $$13\geq \frac{3}{4}y^2$$ since the radicand must be non negative.
2
See OEIS A192629 for
$$1,3,8,120,\frac {777480}{8288641}$$
and OEIS A030063 (which adds $0$ and deletes the fraction) for more information.
-1
Here is the formula for all the solutions of $a^2+b^2=c^2+d^2$
Take $$a=x-y $$
$$b=x^2+xy+y^2+1 $$
$$c=x+y+1 $$
$$d=x^2+xy+y^2$$
For any integers $x$ and $y$
Ex : for $x=2$ and $y=1$ you have
$a=2-1=1$ and $b=2^2+2*1+1^2+1=8$
$c=2+1+1=4$ and $d=2^2+2*1+2^2=7$
And $1^2+8^2=7^2+4^2=65$ the smallest number that can be writing as sum ...
0
Above equation shown below:
$\frac{a^2 + b^2}{ab + 1} = \frac{c^2}{d^2}$ ----$(1)$
For known, $(c,d) =(2,1)$ there is a parametric solution & is shown below:
$a=4wk(2k-1)$
$b=2w(3k-1)(5k-1)$
Where, $w= [(1)/(k^2-4k+1)]$
For $k=4$, we have, $(a,b,c,d)= (112,418,2,1)$
1
After getting $(x^2-x+1)\mid(3x-1)$, and bounding $-2\leq x\leq 3$ as a result, you should first go back and check the condition $(x^2-x+1)\mid(3x-1)$:
$x=-2$: $x^2-x+1=7$, $3x-1=-7\quad\checkmark$
$x=-1$: $x^2-x+1=3$ so doesn't divide $3x-1$.
$x=0,1$: $x^2-x+1=1\quad\checkmark$
$x=2$: $x^2-x+1=3$, so doesn't divide $3x-1$.
$x=3$: $x^2-x+1=7$ does not ...
4
Write $m=x^2-x+1>0$ then from $m\mid 3x-1$ we have $$ 3x\equiv 1 \pmod m$$ and since $m\mid 9m$ we have also $$9x^2-9x+9\equiv 0\pmod m$$ So $$1-3+9\equiv 0 \pmod m \implies m\mid 7\implies m\in \{1,7\}$$
$x^2-x+1 = 1\implies x\in\{0,1\}$
$x^2-x+1 = 7\implies x\in\{-2,3\}$
Check every $x$ and you are done.
2
We have that $$(y^2 + xy)(x^2 - x + 1) = 3x - 1 \implies x^2 - x + 1 \mid 3x - 1$$
$$\implies x^2 - x + 1 \le |3x - 1| \iff x \in [-2, 0] \cup [2 - \sqrt 2, 2 + \sqrt 2]$$
However, $x$ is an integer $\implies x \in \{0, \pm 1, \pm 2, 3\}$
We can set up a table for different values of $x$ and $y^2 + xy$.
$$\begin{matrix} x& -2& -1& 0& 1&...
1
You are on the right track; indeed you have $\gcd(b,c)=2nq\gcd(np,mq)$, and also
$$\gcd(a,d)=\gcd(d,d-a)=\gcd(n^2p^2+m^2q^2+n^2q^2,2n^2q^2),$$
which shows that $\gcd(a,b,c,d)$ divides $2n^2q^2$. It follows that
\begin{eqnarray*}
\gcd(a,b,c,d)&=&\gcd(d,d-a,\gcd(b,c))=\gcd(d,2n^2q^2,2nq\gcd(np,mq)),
\end{eqnarray*}
where in turn
$$\gcd(2n^2q^2,2nq\gcd(...
1
$ax+by=n$ is a line, thus with the density of continuum.
If you take out the double integral (diophantine) solutions, which may be none, or countable (finite or infinite, depending on the bounds), then you are left, at the minimum, with $\mathbb R \backslash \mathbb Z$ (eventually, within the given bounds).
0
If a and b have a common factor the both ax and by have that factor for all x and y so ax+ by has that factor. If n does not have that factor, the Diophantine equation ax+ by= n has no (integer) solutions. For example 2x+ 6y= 5 has no (integer) solutions.
0
It has integer solutions if and only if $\gcd(a,b)\mid n$.
5
Write $$x^2(y-2)^2 = -y^3+3y^2-1$$
Since $y-2\mid -y^3+3y^2-1$ and $y\equiv 2\pmod{y-2}$ we have
$$0\equiv -y^3+3y^2-1 \equiv -8+12-1 \equiv 3 \pmod{y-2}$$ So $$y-2\mid 3\implies y-2\in\{1,-1,3,-3\}$$
so $$y\in\{3,1,5,-1\}$$
Checking each of them we are done.
4
Not suprisingly,
$$x^2y^2 - 4x^2y + y^3 + 4x^2 - 3y^2 + 1 = 0$$
$$\iff (x^2y^2 + y^3 + y^2) - (4x^2y + 4y^2 + 4y) + (4x^2 + 4y + 4) = 3$$
$$\iff y^2(x^2 + y + 1) - 4y(x^2 + y + 1) + 4(x^2 + y + 1) = 3$$
$$\iff (y^2 - 4y + 4)(x^2 + y + 1) = 3 \iff (y - 2)^2(x^2 + y + 1) = 3$$
$$\implies (y - 2)^2 \mid 3 \implies (y - 2)^2 \in \{\pm 1, \pm 3\}$$
Having ...
0
The equation is
$$ax+(a+1)y=a(x+y)+y=az+y=c$$ where $z\ge2,y\ge1$.
Assuming $a\ge0$, There are solutions for all $c\ge2a+1$.
0
My approach solves the equation over the integers, then tries to find a positive solution :
$x=-c$ and $y=c$ is a solution.
as $gcd(a,b)=1$, and $ab-ba=0$, all solutions to the inital equation are of the form $(-c-kb,c+ka)$ for an integer $k$.
assuming $a$ positive, both $x$ and $y$ are positive iff:
$$ \dfrac{-c}{a}< k <\dfrac{-c}{a+1} $$
So a ...
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 $gcd(a, b)$.
0
I'm not certain that I understand exactly what it is you are asking, but there are definitely interesting applications to knowing the ABC conjecture for particular values of $\epsilon, k(\epsilon)$.
To illustrate one such application, suppose that the ABC conjecture holds with $\epsilon=k(\epsilon)=1$. Then for all triples $(a,b,c)$ of relatively prime ...
1
There may be a solution, though one may not have searched hard enough. For example, if your expression with the squares was,
$$D = 5a^2+b^2-c^2-d^2$$
then I found,
$$\left(a+\frac b4+\frac c4+\frac d2\right)\big(4a-b+c-2d\big)-(-a + b + c + d)(a + b - c + d) -\left(-\frac {9b}4+\frac {9c}4-\frac{3d}2\right)\left(b+c+\frac{2d}3\right) = 5a^2+b^2-c^2-d^2$$
...
2
The OP wishes to find more examples of,
$$2b_1^3l_1+3b_1^2l_1^2+b_1l_1^3=2b_2^3l_2+3b_2^2l_2^2+b_2l_2^3=k\tag1$$
or equivalently,
$$p q (p + q) (2 p + q) = r s (r + s) (2 r + s)=k\tag2$$
One solution to this is,
$$p,q = 3,4\\ r,s = 5,2$$
with $k=840$ and which obviously has the auxiliary relation $p+q = r+s$. So let,
$$p,\;q = a + b + c,\; -a - b + c\\...
0
Note that the given equation is $$x^3+y^3+3xy=6021$$ or $$x^3+y^3-1+3xy=6020$$
Factoring it we get , $$(x+y-1)(x^2+y^2+1+x+y-xy)=2^2.5.7.43$$
Obviously check $\equiv 3$ and see $$x+y-1\equiv 2 \mod 3$$
Also, $$x+y-1 < x^2+y^2+1+x+y-xy$$
This means, $$x+y-1 \rightarrow 20,5,2,35$$
so now it's easy to see that $$x+y-1=20$$ and so $(x, y)=(18,3)$ or $(3,...
0
The OP states
... but, I found that solution by solving the linear diophantine equation
From this site I found
Here the equation is $7x\equiv 41 \pmod {13}$ or
$$\tag 1 7x\equiv 2 \pmod {13}$$
Here also, multiplication $7 \times x$ is repeated addition. So you can also get the inverse of ${[7]_{13}}^{-1}$ and/or the solution of $\text{(1)}$ using the ...
1
If you have $x^2+2y^2=8z+5$, by looking $\bmod 2$, you have that $x$ has to be odd.
Now (a very useful fact for me) we have that $odd^2\equiv 1\mod 8$. In particular $x^2\equiv 1\mod 8$.
So, if you look your original equation $\bmod 8$ you get $$1+2y^2=5\bmod 8$$
This implies that $2y^2\equiv 4\mod 8$ and by dividing by $2$, you get $y^2\equiv 2\mod 4$. ...
7
If you look at the equation mod $8$, note that squares are either $0,1$ or $4$ mod 8. Thus the left hand side can only attain the values $0,1,2,3,4$ or $6$ mod $8$ and hence will never be equal to the right hand side, which is $5$ mod $8$.
1
Alternatively, you may write $$2x - 6y = 2(x-3y).$$ Since $x$ and $y$ are integers, so must be $x-3y$. So $2(x-3y)$ must be an even integer, clearly being divisible by $2$. But $3$ is odd.
3
Rewind to the point where you say $x=3y+3/2$. We rearrange this to $x-3y=3/2$, then note that since we have taken $x$ and $y$ to be integers, $x-3y$ is also an integer. But $3/2$ is not an integer, a contradiction.
3
Jorn Steuding, Diophantine Analysis.
Ed Burger, Exploring the Number Jungle: A Journey into Diophantine Analysis.
Ed Burger and Robert Tubbs, Making Transcendence Transparent.
Nigel Smart, The Algorithmic Resolution of Diophantine Equations.
1
Yes, the five complex roots are $\omega^k \sqrt[5]{2}$, where $\omega=\exp(2\pi i/5)$ and $k=0,\dots,4$.
0
Hint:
The equations can be seen as the dot and cross products of three vectors formed by the common origin $(a,b)$ and three points.
The ratios $\dfrac{\text{det}}{\text{dot}}$ give you the (tangent of the) angles they form, and the square roots of the pairwise sum-of-squares give you the products of the lengths. From the latter you can retrieve the ...
0
I will just give a sketch for now how to solve it, I can fill in more details if necessary.
First define vectors $v_i = (x_i-a,y_i-b),\, i = 1,2,3$. Then we can rewrite the equations as
\begin{align}
\langle v_1, v_2\rangle = a_{12},\ \langle v_2, v_3\rangle = a_{23},\ \langle v_3, v_1\rangle = a_{31},\\
\det[v_1^t\ v_2^t] = b_{12},\ \det[v_2^t\ v_3^t] = ...
1
I think, the way with using Pythagorean triplets is the best.
Let $\gcd(a,b,c)=1$.
Thus, there are natural $m$ and $n$ with different parity such that $m>n$ and $\gcd(m,n)=1$ and
$a=2mn,$ $b=m^2-n^2$.
Thus, $c^2=m^2+m^2$ and by the same way there are naturals $p$ and $q$ with a different parity, such that $p>q$, $\gcd(p,q)=1$ and $m=2pq$ and $n=p^2-...
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