All Questions
51
questions
20
votes
4
answers
578
views
When can products of linear terms differ by a constant?
We have
$$ X(X+3) + 2 = (X+1)(X+2)$$
and
$$ X(X+4)(X+5) + 12 = (X+1)(X+2)(X+6)$$
and
$$ X(X+4)(X+7)(X+11) + 180 = (X+1)(X+2)(X+9)(X+10).$$
Do similar polynomial identities exist for each degree?
That ...
- 2,071
0
votes
0
answers
54
views
Simplest failures of Hasse principle
I’m trying to understand the simplest cases where a Diophantine equation has solutions in the reals and modulo all prime powers, but not in the integers.
I know classes of examples of degree 6 in 1 ...
- 361
1
vote
1
answer
39
views
Can relative (or even absolute) quotient size be calculated from a list of polynomials which are multiples of a given variable?
I am working on a Diophantine equation in integers $x$ and $y$. The equation has been solved, so I already know the solutions (there are four) — I am trying to find a more elementary solution.
Through ...
- 7,929
0
votes
1
answer
80
views
Parametric solution of quartic diophantine equation in three variables
How can I handle the quartic diophantine equation in three variables $x$, $y$ and $z$ $$x^4-x^2=y^2-z^2$$ in general, i.e, does exists a (three-variable) parametric solution?
What I've tried is ...
3
votes
2
answers
421
views
Polynomial system of equations over integers
I want to solve the system of equations:
$$\begin{cases}
x^4+4y^3+6x^2+4y = -137 \\
y^4+4x^3+6y^2+4x = 472
\end{cases}
$$
$x, y \in \Bbb{Z}$.
It most definitely amounts to messing around with algebra ...
- 1,488
2
votes
1
answer
202
views
Solving a degree-6 Diophantine inequality
While working on a proof, I ended up with the following Diophantine inequality of degree-6:
$$
a_1^6-2a_1^5+a_1^4 (1-4n)+8a_1^3 n+a_1^2 (4n^2+8n)-8a_1 n^2+16n^2 \ge 0 \tag 1
$$
The variable is $a_1$ ...
- 3,341
4
votes
2
answers
153
views
determine if the equation $x^n+y^n+z^n+w^n=u^{n+1}$ has infinitely many solutions in distinct integers
Let $n\ge 1$. Determine if the equation $x^n+y^n+z^n+w^n=u^{n+1}$ has infinitely many solutions in distinct integers. If so, determine if there are two solutions $(x_i,y_i,z_i,w_i,u_i)$ for $i=1,2$ so ...
- 1,166
2
votes
4
answers
208
views
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
My try: I think $(x;y)=(0;2)$ is only solution. So I try prove $y\ge3$ has no solution, by $(x+1)(x+6)-2=2^y$.
So $2\mid (x+1)(x+6)$, but this is ...
- 727
2
votes
2
answers
58
views
Factorizations versus integer solutions in single-variable quartics
If $x$ is a positive integer, and I have the factorization
$$(x-1)(79x^3+159x^2-513x+255)=0,$$
what is the easiest way to conclude that $x=1$ is the only integer solution?
Related question: If [in a ...
- 7,929
2
votes
2
answers
135
views
Perfect Square With Two Integer Variables
I am trying to solve a number theory problem in general form. However, I got stuck in the following step:
$a,b,n \in \mathbb Z^{+}$ for which values of $n$, this equation is solvable $\frac{(n+1)(n+2a)...
- 49
0
votes
3
answers
272
views
Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.
Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.
I found a solution, but it involves a lot of case work. Can someone help me find a solution which doesn't involve a lot of ...
- 255
1
vote
3
answers
204
views
Cubic diophantine equation in two variables
I am trying to solve the Diophantine equation $xy^2 + 2xy + x - 243y = 0$.
I simplified it to $x(y^2 +2y +1) = 243y$ but I am stuck on what to do now. Any help would be appreciated.
- 59
2
votes
1
answer
227
views
When does $(xz+1)(yz+1)=az^{3}+1 $ have finitely many solutions in positive integers?
Consider the diophantine equation in three variables $x$, $y$ and $z$; ($xz+1$)($yz+1$) $=$ $6z^{3}+1$. The only positive integer solutions I have found are {$x=4,y=10,z=7$} and {$x=10,y=4,z=7$}. From ...
- 234
1
vote
5
answers
230
views
Quadratic Diophantine equation $x^2+6y^2-xy=47$ has no solutions.
I am trying to show that $x^2 + 6y^2 - xy = 47$ has no integer solutions. I know that the an efficient way is to look at this equation modulo $n$; other equations can be easily be solved this way. I ...
- 854
0
votes
2
answers
116
views
Solving the Diophantine equation $k^2(k+1)=m(3m-1)$
I am working on solving the following Diophantine equation:
$$k^2(k+1)=m(3m-1)$$
And so far I solved, using WolframAlpha, the following solutions:
$$(k,m)=(-1,0);(0,0);(1,1);(4,-5);(6,-9)$$
Is ...
- 109