TITLE: How find this equation all solution $2a^2-1=b^{2013}$
QUESTION [8 upvotes]: let $a,b\in \mathbb Z$, and such $$2a^2-1=b^{2013}$$
find all value $a,b$
I think $(a,b)=(0,-1),(1,1),(-1,1)$ is solution, and have other solution.  Thank you everyone

REPLY [3 votes]: Following Jyrki's suggestion, set $c=b^{671}$ and the equation becomes $$2a^2=c^3+1$$
Multiply both sides by 8 to get $$(4a)^2=(2c)^3+8$$
This is a Mordell curve, which have finitely many solutions.  In this case the constant is $8$, and here we can find the only integral solutions $$(4a,2c)\in \{(0,-2),(3,1),(-3,1), (4,2), (-4,2), (312,46), (-312,46)\}$$
This gives $$(a,c)\in \{(0,-1),(1,1),(-1,1),(78,23),(-78,23)\}$$
However only the first three give $c$ that is a 671st power.  This confirms that there are three solutions in the integers.

Followup: We may replace $2013$ with $3(2k+1)$ for any integral $k\ge 2$ and get the same result.