Each of
a,
b, and
c is a positive integer satisfying this relationship:
a2+b2+1=c2
Is ⌊
a/2⌋+ ⌊
b/2⌋
always even?
If so, prove it.
If not, provide a counter example.
Note: ⌊n⌋ denotes floor(n), that is, the greatest integer less than or equal to n.
Write a^2+b^2=(c^2-1), and assume a,b, both odd, so that a=(2x-1), b=(2y-1)
Then a^2+b^2 = (2x-1)^2+(2y-1)^2 = 2(2x^2-2x+2y^2-2y+1), which is singly even.
Hence this expression is worth 2, mod4, making c^2 worth 3, mod4. However there are no such squares.
Now assume one of a,b is even, so that, say, a=(2x-1), b=(2y).
Then (2x-1)^2+(2y)^2 = 4x^2-4x+4y^2+1, which is clearly worth 1, mod4, making c^2 worth 2, mod4. However there are no such squares.
So both a and b are even, and c is an odd square, say (2z-1)^2: (2x)^2+(2y)^2=(2z-1)^2-1
Since a = (2x), then ⌊a/2⌋ is simply x; since b = (2y), then ⌊b/2⌋ is simply y. If x and y are themselves even, then (x+y) is even, similarly if both are odd, and we are done.
Assume however that one of (x,y) is even, say, x=2p and the other odd, say, y=(2q-1).
Now we have (2(2p))^2+(2(2q-1))^2=(2z-1)^2-1
Expanding: 16p^2+4(2q-1)^2 = (2z-1)^2-1
Solve for z: z = 1/2(1-sqrt(16p^2+16q^2-16q+5)), which is never positive, or z = 1/2(sqrt(16p^2+16q^2-16q+5)+1).
Since 1/2 of 1 = 1/2, it follows that sqrt(16p^2+16q^2-16q+5) also contains the same fraction, and so is rational and some odd multiple of 1/2.
Let sqrt(16p^2+16q^2-16q+5) = (2w-1)/2, for some positive integer, w, to produce the required odd 1/2.
Squaring both sides: (16p^2+16q^2-16q+5) = ((2w-1)/2)^2, when w = 1/2(2sqrt(16p^2+16q^2-16q+5)+1)
Now z is an integer and w is an integer, but w-2z = 1/2(2sqrt(16p^2+16q^2-16q+5)+1)-2*1/2(sqrt(16p^2+16q^2-16q+5)+1) = -1/2, a contradiction.
So the expression ⌊a/2⌋+ ⌊b/2⌋ is always even.
Edited on November 18, 2022, 7:23 am
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Posted by broll
on 2022-11-18 07:19:41 |
Possible Solution
