Question 1106492: m^2+n^2=-mn
Can this expression be true for any value of (m,n)?
Given (m,n) not equal to 0.
Answer by ikleyn(53765)   (Show Source):
You can put this solution on YOUR website! .
No.
Counter-example: m = 1, n = 1.
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By the way, what we see in the post, is NOT expression.
It is an equation, or an equality, or an (hypothetical) identity . . .
/\/\/\/\/\/\/\/\/\/
I suspect that the formulation in the post is NOT that you wanted to know/to ask.
More meaningful questions are:
1) Can this equality be true for all pairs (m,n) of integer numbers?
2) Can this equality be true for some pairs (m,n) of integer numbers? At least for one pair ?
(providing m=/= 0, n=/= 0 in these pairs (in this pair) ).
Solution
1) Can this equality be true for all pairs (m,n) of integer numbers?
I just answered this question above by giving a counter-example: No, it can not.
2) Can this equality be true for some pairs (m,n) of integer numbers?
m^2 + n^2 = -mn.
Again, the answer is NO.
a) if m and n are both odd numbers, then the left side is an even integer number, while the right side is an odd integer.
Contradiction.
b) if m is an even integer and n is an odd, then the left side is an odd, while the right side is an even.
Contradiction.
c) if m is an odd integer and n is an even, then we have the same contradiction.
d) If both m and n are even, then in the given equality we can reduce both m and n by dividing each by 2, and we will repeat it
until we get one of the three cases a), b) or c).
Then exactly as in each of these cases we will get the same contradiction.
The problem is solved.
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