Are there any integer solutions to this equation? Anyway, suppose what they want is not true. Of course, 2 divides $y+1$ and if we write the equation as $ ((y^5+1)/2)( y^{10}-y^5+1)=x^2$ we observe that $ ((y^5+1)/2, y^{10}-y^5+1)$ is a divisor of 3. If it is 1, then $y^{10}-y^5+1$ is a perfect square, impossible for y at least 2 (^we can put it between two consecutive squares). So, suppose it is 3. Then we can write $ y^5+1=6A^2$ and $y^{10}-y^5+1=3B^2$ with $x=3AB$. Since $ y^5-y$ is a multiple of 3, it follows that $ y+1$ is a multiple of 6. Thus, if we write $ ((y+1)/6)(y^4-y^3+y^2-y+1)=A^2$ we see that since 5 does not divide $A$, we have $ ((y+1)/6, y^4-y^3+y^2-y+1)=1$ and thus $ y^4-y^3+y^2-y+1$ is a square. Again, this is impossible by the fact that $ (2y^2-y)^2<4(y^4-y^3+y^2-y+1)<(2y^2-y+2)^2$.

See here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=399294&hilit=do+there+exists+marathon No solution exists..

Since the statement of the original question is false (as pointed out by there having no integer solutions), arent such claims such as $5|x$ logically true ?

ck204 wrote: See here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=399294&hilit=do+there+exists+marathon No solution exists.. Sorry but the solution is not true....

by mod 5 $2x^2 -1 \equiv y^3 \mod 5$ $x^2 \equiv (y^3+1)*-2 \mod 5$ notice (-2) is not quadratic residue mod 5 so does $(y^3+1)$ so $(y^3+1)^2 \equiv 4 \mod 5$ $(y^3+1) \equiv 2,-2\mod 5$ $y^3 \equiv 1 ,-3 \mod 5$ $5 \nmid x$