I've been looking at this IMO 2006 contest exercise and found a thing I cannot understand:
The exercise is:
Now what I do not understand is , why
???
Can you explain me this? Thank you in advance!
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First, go back to the original equation. Clearly if $x<-1$ the LHS is not an integer, if $x=-1$ the LHS is not a square, and if $x=0$ we have the solution $y=\pm2$. So now assume $x>0$.
As $y-1$ and $y+1$ differ by $2$, they are both even or both odd. They can't be both odd since their product is even, so they are both even. In fact, they are consecutive even numbers and this is why one of them is a multiple of $4$ and the other isn't.
This shows that the RHS, and hence also the LHS, is a multiple of $8$, and this is why $x\ge3$.
So $y-1$ and $y+1$ have between them exactly $x$ factors of $2$. One of them is even but not a multiple of $4$: that is to say, it has exactly one factor of $2$, and the other has the remaining $x-1$ factors. Thus either $y-1$ is $2^{x-1}$ times an odd number $m$, which gives $$y=2^{x-1}m+1\ ,$$ or $y+1$ has this property, which gives $$y=2^{x-1}m-1\ .$$ The last equation in your post combines these two equations into one.
David
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