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determine whether the property is true???

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2 Answers

Daniel,
 
I'm not sure what the question marks represent in your problem statement ?(x+y)?^2 = x^2 + y^2 , so I will ignore them in my solution.  If this is incorrect, please restate the problem for me.  Thanks.
 
So the relation would be (x + y)2 = x2 + y2 if the question marks are removed.   If we are talking about integers:  .. -2, -1, 0, 1, 2, ...   then 
 
by cross-multiplying we get x2 + 2xy + y2 = x2 + y2, which would mean 2xy = 0, so x = 0 or y = 0.  So the statement is true if either x or y is equal to zero, and otherwise it's false.  
 
But let's explore this a bit further.
 
Suppose you were working on a number system that was like the integers but not exactly - say on a 6-hour clock.  So that, for example,  2*3 = 6 which is equal to 0 on a 6-hour clock.  So on this 6-hour clock, if x = 2 and y = 3, then xy = 0 and we would have (x + y)2 = x2 + ywhere neither x nor y was zero.  So it really depends which number system you are working on whether an equality like this holds.  
 
 
 
 This problem can be approached geometrically:
 
      a^2 + b^2 = c^2
 
      this relationship is Pythagoras theorem of a right triangle with legs sizes a,b and hypotenuse size c
 
     Taking square of both sides.
 
 
     √( a^2 + b^2) = c
 
         now, if we let  √a^2 + b^2 equals  a+ b , we are violating the Pythagoras law, implying that
        
            length of the hypotenuse equals sum of the 2 length, which is not true.
 
           Summation of squares being equal square of some quantity does not imply that summation of
 
           length is equal
 
              √ ( 4^2 + 3^2 ) = √( 16 +9 ) = √25 = 5
 
             √4^2 + √3^2 = 4 + 3 = 7
 
              √( a^2 + b^2) ≠ a +b, then
 
              a^2 + b^2 ≠ ( a + b) ^2