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bio website math.ubc.ca/~bennett
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visits member for 1 year, 8 months
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Sep
9
comment Finding every triplet $(n,a,b)$ such that $n!=2^a-2^b$
Ummm, if $r$ is a primitive root mod $p^2$, then $r$ is a primitive root mod $p^k$ for all $k$. Luck is not involved…
Sep
6
comment Finding every triplet $(n,a,b)$ such that $n!=2^a-2^b$
$2$ is a primitive root modulo $9$ and hence modulo $3^k$ for all $k$.
Sep
5
answered Finding every triplet $(n,a,b)$ such that $n!=2^a-2^b$
Sep
4
answered On some inequalities which are an generalization of F. Beukers' corresponding results
Aug
2
comment How to prove $~(c - b) ^ 2 + 3cb = x^3~$ has no nonzero integer solutions?
If, say, $c=1, b=-1$ and $x=1$, we have a solution to (2).
Jun
26
comment How to find the minimum value of $|5^{4m+3}-n^2 |$
The minimum is indeed $275$ but I do not see a short, mathematics-free way to prove this….
May
9
comment Find all real $x$ ,such $8x^3-20$ and $2x^5-2$ is perfect square
Are you really offering a bounty for a problem whose solution is in the comments?
May
9
comment Infinitely many perfect squares
That doesn't affect the genus. Just multiply the equation by $(A 2^{n_0})^2$ to get a new equation of the shape $Y^2=X^3+C$ with $C=(A 2^{n_0})^2 B$.
May
7
comment Find all real $x$ ,such $8x^3-20$ and $2x^5-2$ is perfect square
Plugging f:=EllipticCurve([0,-20]); followed by IntegralPoints(f); into Magma ensures that the solution Robert found is the only one (i.e. $x=3$).
May
7
answered Infinitely many perfect squares
May
2
awarded  Nice Answer
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.