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1d
comment Integers points of an elliptic curve
If you assume your curve to be given by a minimal model or some such (to avoid simply scaling rational points to give, supposing positive rank, as many integral points as desired), the current belief is that the number of "integral" points should be absolutely bounded.
Apr
21
awarded  Editor
Apr
21
comment Finding integer solutions of $m^2-n^5 = m - n$
There is no reason to believe that there is an elementary approach to solve this problem. Sometimes Diophantine equations are just hard.
Apr
21
revised Finding integer solutions of $m^2-n^5 = m - n$
added 1 character in body
Apr
21
answered Finding integer solutions of $m^2-n^5 = m - n$
Apr
15
answered Diophantine equation resembling FLT
Mar
26
comment diophantine-equations
There are infinitely many solutions (in parametrized families) for $k=3, 4$ and $5$ (see a paper of Jonny Edwards in Crelle's journal), finitely many for $k=6$ (well, none with $x$ and $y$ positive, I suspect) and probably not many others with $k \geq 7$ (hard to prove, I also suspect).
Mar
3
answered Is $53$ expressible in this form?
Jan
23
awarded  Yearling
Dec
24
comment Does $p_{1}^x + p_{2}^y = n$ have uniqe solution for $x$ and $y$ ($p_{1}, p_{2}$ are primes).
Not to be unkind, but both the "proof' and the accepted answer here are incorrect.
Dec
19
answered Does $p_{1}^x + p_{2}^y = n$ have uniqe solution for $x$ and $y$ ($p_{1}, p_{2}$ are primes).
Dec
19
comment Does $p_{1}^x + p_{2}^y = n$ have uniqe solution for $x$ and $y$ ($p_{1}, p_{2}$ are primes).
How about $2^3+3=2+3^2$? The conclusion that $p \equiv 1 \mod{q}$ and $q \equiv 1 \mod{p}$ does not follow
Nov
29
answered Approximate irrational numbers with the same denominator
Nov
23
answered Solve $3^a-5^b=2$ for integers a and b.
Oct
23
comment Integral solutions of $x^5-27y^3=2x$
What is your motivation for studying this?
Oct
4
awarded  Mortarboard
Sep
25
answered Is there any infinite set of primes for which membership can be decided quickly?
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 $n!=2^a-2^b$