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Rooted to the Floor (Posted on 2013-09-24) Difficulty: 3 of 5
Find all pairs (x,y) of integers satisfying:

floor(√(2x)) = floor(y*(2 + √2))

No Solution Yet Submitted by K Sengupta    
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anaylitical solution Comment 1 of 1

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
k-1<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
a-1<k/z<=a<=(k+1)/z<a+1
thus k must be in the intersection of the two intervals
((a-1)z,az] and [az-1,(a+1)z-1)
since az-1>(a-1)z and az>(a+1)z-1 for a>1 then
this is the same as the interval
[az-1,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 2013-09-25 11:54:01
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