let f(x)=floor(x) and c(x)=ceiling(x)
if f(x)=k for some integer k then we have
k<=x<k+1
also if c(x)=k for some integer k then we have
k1<x<=k
for compactness let z=2+sqr(2)
f(sqr(2x))=f(yz)
if f(sqr(2x))=f(yz)=k for some integer k then we have
k<=sqr(2x)<k+1
k^2<=2x<(k+1)^2
thus x can be any even integer in the interval [k^2,(k+1)^2)
there is an even integer in this interval for all k>=1
we also have k<=yz<k+1
(k/z)<=y<(k+1)/z
thus y can be any integer in the interval [k/z,(k+1)/z)
this interval contains an integer only when
c(k/z)<=f((k+1)/z)
since the length of this interval is a constant 1/z<1
this inequality only holds at equality thus we need
c(k/z)=f((k+1)/z)=a for some integer a
this gives us
a1<k/z<=a<=(k+1)/z<a+1
thus k must be in the intersection of the two intervals
((a1)z,az] and [az1,(a+1)z1)
since az1>(a1)z and az>(a+1)z1 for a>1 then
this is the same as the interval
[az1,az]
the only integer in this interval is f(az)
thus y can be any integer in the interval [f(az)/z,(f(az)+1)/z)
so all solutions are given by:
z is 2+sqr(2)
a is an integer greater than 1
x is an even integer in [f(az)^2,(f(az)+1)^2)
y is an integer in [f(az)/z,(f(az)+1)/z)

Posted by Daniel
on 20130925 11:54:01 