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bio website math.ubc.ca/~bennett
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visits member for 1 year, 2 months
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Jan
23
awarded  Yearling
Dec
17
answered How find this equation $n!+(n+1)!+\cdots+(n+m)!=a^b$ all solutions
Nov
26
comment There are only finitely many integer solutions to $ax^n+by^n=c.$
Well, $ax^3+by^3=cz^3$ can certainly have infinitely many coprime integral solutions...
Nov
22
comment What integers can be represented by the quadratic form $4x^2 - 3y^2 - z^2$?
This case is rather easier as the form is indefinite (and can be easily shown to represent every $n \not\equiv 2 \mod{4}$ infinitely often).
Nov
21
comment What integers can be represented by the quadratic form $4x^2 - 3y^2 - z^2$?
Did you perhaps look for solutions? There are many.
Nov
20
comment Show that there are no squares included in the sequences
And $3$ is not a quadratic residue modulo $5$, for the final case.
Nov
6
comment Generalization of an inequality $0\lt e^6-{\pi}^4-{\pi}^5\lt 0.00002$
I mean $l^{-1}$ instead of $a^{-1}$ in the above comment.
Nov
6
comment Generalization of an inequality $0\lt e^6-{\pi}^4-{\pi}^5\lt 0.00002$
If we have even $| e^k - \pi^l - \pi^m| <1$ with, say, $l > m$, then, $\left| \log \pi - k/l \right| \ll a^{-1} \pi^{m-l}$ and so, if $l$ is a fair bit bigger than $m$, we would have an extraordinarily good approximation to $\log \pi$. Unfortunately, as far as I know, we cannot rule this out, but one suspects that such an approximation is unlikely to exist. This would suggest that $|l-m| \ll \log m$. For such pairs $(l,m)$, a heuristic argument suggests that we cannot force $e^k-\pi^l-\pi^m$ to be arbitrarily close to $0$.
Oct
24
comment How does one attack a divisibility problem like $(a+b)^2 \mid (2a^3+6a^2b+1)$?
While the original question can be reduced to finding the integral points on a parametrized family of elliptic curves (certainly not just one), it's extremely unclear that his approach is even slightly helpful for actually solving the problem.
Oct
24
answered Do there exist natural number solutions such that $x^m=11\cdots11$ for $m\ge 2$?
Oct
16
comment About the integer solutions of $\ y^2=x^3+9$
There might not be an easy proof. The most straightforward that I know uses Skolem's method -- for this equation, this was done in a thesis of Hemer from 1952. Your equation reduces to solving $Aa^3-Bb^3=C$ for all triples $A, B, C$ with $ABC=6$. I don't know an elementary way to do this.
Oct
15
comment About the integer solutions of $\ y^2=x^3+9$
These are the only integer solutions. Data for $y^2=x^3+k$ with $|k| < 10^4$ can be found at tnt.math.se.tmu.ac.jp/simath/MORDELL. These results were proved using lower bounds for linear forms in elliptic logarithms (though the case $k=9$ can be treated more simply, by reducing to cubic Thue equations).
Sep
17
comment Does this Diophantine cubic have solutions?
An old theorem of Delone and Nagell (discovered independently) says that, given $a$ and $b$, the equation $ax^3-by^3=1$ has at most one solution in positive integers $x$ and $y$ (and moreover tells you how to find the solution, if it exists).
Aug
24
comment What is the smallest real $q$ such that there is always a prime between $n^q$ and $(n+1)^q?$
What do you mean by ``prime distance is $O(\log n)$''? The best known result on largest prime gaps would give the desired result for a given $q > 40/19$, for suitably large $n$ (Baker, Harman and Pintz).
Aug
5
comment A natural number equation
Precisely correct. And, for $m > 3$ (even or odd), this equation has no solutions...
Aug
5
comment A natural number equation
The case $n=7$ demonstrates the lack of sufficiency. If one just considers the discriminant of the original equation with respect to $y$, one is led to Pell equations without solutions, provided $n>1$.
Jul
24
answered How many solutions are possible to the equation $a^x-b^y=c$?
Jul
23
comment Erdös-Straus conjecture
Mordell's book "Diophantine Equations" has a section on this.
Jul
17
comment Is there any pythagorean triple (a,b,c) such that $a^2 \equiv 1 \bmod b^{2}$
Yup, this looks like it follows from an old paper of mine with Gary Walsh in Indagationes Math. from 2000.
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.