Find
all integer solutions of:
x2+y2+
z2=2x-3z+5
x2+y2+3x=2
Any remarks?
Given
x^2+y^2+z^2=2x-3z+5
x^2+y^2+3x=2
3(x^2+y^2+z^2) = 6x-9z+15 Multiply by 3
2x^2+6x+2y^2 = 4 Multiply by 2
3(x^2+y^2+z^2)-6x+2x^2+6x+2y^2 = 15-9z+4 Adding
5x^2+5y^2+3z^2=19-9z Simplifying
5(x^2+y^2)=19-9z-3z^2 Consolidate terms in z.
Note that:
19-9z-3z^2=-3(z+1)(z+2)+25
Then
5(x^2+y^2)-25= -3(z+1)(z+2) Rearranging
(x^2+y^2)-5= -3/5(z+1)(z+2) Divide through.
Everything on LHS is integer, but there is a fraction on RHS.
Remark, however, that the puzzle requires 'No fractions'.
So (x^2+y^2)=5, {1,2}{-1,2}{1,-2}{-1,-2} and vice versa, and z must be either -1 or -2, for a total of 16 solutions.
Edited on July 14, 2022, 7:47 am
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Posted by broll
on 2022-07-14 07:41:19 |
No Subject
