meta user
network profile
| bio | website | math.ubc.ca/~bennett |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 4 months |
| seen | 5 hours ago | |
| stats | profile views | 40 |
|
May 19 |
comment |
A binary quadratic form: $nx^2-y^2=2$ The $n=2p$ case follows from an old result of Nagell; it does not depend upon the class number of the corresponding fields. |
|
May 15 |
answered | About the infinite solutions of a Diophantine equation |
|
Apr 23 |
answered | Integer solutions of $x^3+y^3=z^2$ |
|
Apr 23 |
answered | Integer solutions of $n^3 = p^2 - p - 1$ |
|
Mar 28 |
comment |
Minimum of $n$? $123456789x^2 - 987654321y^2 =n$ ($x$,$y$ and $n$ are positive integers) I'm running around a bit today, but if we have $$ax^2-by^2=c$$ with corresponding ``large'' partial quotient $a_n$ to $\sqrt{b/a}$, we should have something like $a_n |c|=2 \sqrt{ab}$ (where the error here can be nicely bounded). For the example here, with the largest partial quotient noted, we have $$2 \sqrt{ab}/a_n =225.000077 \cdots$$ which is suggestive. |
|
Mar 28 |
awarded | Commentator |
|
Mar 28 |
comment |
Minimum of $n$? $123456789x^2 - 987654321y^2 =n$ ($x$,$y$ and $n$ are positive integers) One uses $$\frac{1}{q_n (q_{n+1}+q_n)} < \left| \theta - \frac{p_n}{q_n} \right| < \frac{1}{q_n q_{n+1}}$$, the recurrence for the $q_i$, the values you know for the partial quotients and the fact that one can locally exclude most of the values corresponding to $123456789p^2−987654321q^2$. |
|
Mar 28 |
comment |
Minimum of $n$? $123456789x^2 - 987654321y^2 =n$ ($x$,$y$ and $n$ are positive integers) And as an additional hint, you might note that the $18577$th partial quotient is $1410862$, corresponding to a convergent $p/q$ for which $123456789p^2-987654321q^2=495$. From basic Diophantine approximation, it is easy to figure out what the $a/b$ corresponding to the larger partial quotient David notes leads to when you evaluate $123456789a^2-987654321b^2$, without actually doing any further computation. |
|
Mar 12 |
comment |
Solve for diophantine equation $x^n + y^n + z^n =1$ More generally (for $n$ odd), it's unknown if there are any nontrivial solutions to $x^n+y^n=z^n+w^n$, for $n \geq 5$. |
|
Mar 1 |
answered | Solve $a^3-5a+7=3^b$ over the positive integer |
|
Mar 1 |
comment |
Solve $a^3-5a+7=3^b$ over the positive integer I'm not sure how one could make this induction work (it might, but controlling sizes in Hensel lifts can be rather delicate). |
|
Feb 24 |
awarded | Enthusiast |
|
Feb 17 |
comment |
Least power. Squares again Well, $2x^2+1$ is a square for infinitely many values of $x$, for example, but if one checks to see that the equation $y^2=f(x)$ corresponds to a curve of positive genus, then there are at most finitely many such solutions, via a theorem of Siegel. This is certainly true, for instance, if $f(x)$ is irreducible of degree at least $3$. I believe a classification of $f$ for which the corresponding curve has genus $0$ was carried out by Bilu and Tichy. |
|
Feb 13 |
comment |
Diophantine equations - Perfect square and Perfect cube related Next step : which primes can divide $y^2+4$? |
|
Feb 12 |
comment |
Diophantine equations - Perfect square and Perfect cube related As a hint for Problem 1), try adding 4 to both sides of the equation and see what happens... |
|
Feb 8 |
comment |
Find $x$ such that $12+13^x$ be a perfect square Thank you, Eric. |
|
Feb 7 |
answered | Find $x$ such that $12+13^x$ be a perfect square |
|
Jan 31 |
comment |
Discriminant of isogenous elliptic curves Other than the fact that they are both divisible by the conductor (and presumably satisfy Szpiro's conjecture), I think this is quite subtle. Even the case of conductor $30$ reveals complications.... |
|
Jan 30 |
answered | Prime divisors of $n^{3} - 27$ and squarefreeness |
|
Jan 29 |
answered | $a^m+k=b^n$ Finite or infinite solutions? |
