Find
all integer solutions of:
x2+y2+
z2=2x-3z+5
x2+y2+3x=2
Any remarks?
(In reply to
Puzzle Answer by K Sengupta)
Multiplying each of the given equations by 4, we have:
4x^2 -8x+y^2+4z^2+12z=20
4x^2+12x+4y^2=8
Let (2x-2, 2z+3)=(P, Q) (say)
Then, we must have:
P^2+4y^2+Q^2= 33 ..... (i)
(P+5)^2 +4y^2 =17 .....(ii)
From (ii), we must have:
(P+5, y) = (1,2), (1, -2), (-1,2), (-1,-2)
=> (P, y) = (-4, 2), (-4, -2), (-6, 2), (-6, -2)
Since |P| =6 forces P^2 > 36 > 33, it is untenable.
So, (P, y) =(-4, 2), (-4, -2)
Then from (i), we obtain:
(P, y, Q) = (-4, 2, 1), (-4, 2, -1), (-4, -2, 1), (-4, -2, -1)
=> (x, y, z) = (-1, 2, -1), (-1, 2, -2), (-1, -2, -1), (-1, -2, -2)
Explanation to Puzzle Answer
