Mike Bennett
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 Nov 17 comment Original proof of Ljunngren's equation The original proof uses a very delicate variant of Skolem's $p$-adic method. It can probably be recast as an application of Chabauty-type arguments, for what that's worth. Sep 8 comment Equations system question Well, and $(x,y)=(2,\pm1)$. Sep 6 comment Positive integer solutions to $a^x-b^y=2$ This is the only known solution. It is an open problem as to whether it is the only solution. Aug 28 comment Thue equation $x^4 - 6 x^3 y - x^2 y^2 + 6 x y^3 - y^4=-1$ In Pari/GP (which is freely available and documented) you can type : thue(thueinit(x^4-6*x^3-x^2+6*x-1),-1) to get a reply of [[-1, 1], [1, -1], [0, 1], [0, -1], [1, 1], [-1, -1], [15, 17], [-15, -17], [6, 1], [-6, -1]]. This "proves" that the solutions are as listed. The Thue solver here depends upon lower bounds for linear forms in (complex) logarithms and lattice basis reduction (facts which may or may not interest you). Aug 23 comment Does $(1^a+2^a+3^a+4^a+5^a)^b=1^c+2^c+3^c+4^c+5^c$ imply $(a,b,c)=(1,2,3)$? I don't believe that the OP is looking for a polynomial identity (it is easy to show that there are no others). I suspect he (or perhaps she) simply wants to rule out the existence of quadruples $(a,b,c,n)$ for which $(\sum_{k=1}^n k^a)^b=\sum_{k=1}^n k^c$ with $(a,b,c) \neq (1,2,3)$, a rather subtler problem. Aug 22 comment How to solve special type of Diophantine equation The integer solutions to this equation have been determined by Bugeaud, Mignotte, Siksek, Stoll and Tengely : see arxiv.org/pdf/0801.4459v4.pdf Jul 27 answered Generalization of Erdos-Selfridge May 2 comment Only finitely many $a, b$ such that $2+3^n+5^{n^2}=2^a7^b$ for some $n$? There are no solutions modulo $24$. Apr 25 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 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