2 5 9 2 a 2 + + 3 1 9 2 b 2 = = 1 6 8 8 4 2 8 4 4 2 1
Assuming a and b are positive integers, find a + b .
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Is this representation unique? Why?
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A very good question... of course, given the fact that Brilliant only accepts one answer, it should be unique.
I checked.
8 4 4 2 1 is a prime of the form p = 4 n + 1 . That guarantees that there exists a unique representation as the sum of squares.
By the Fibonacci-Brahmagupta Identity,
( a 2 + b 2 ) ( c 2 + d 2 ) = ( a c + b d ) 2 + ( b c − a d ) 2
Plugging in c = d = 1 , we get
( a 2 + b 2 ) ( 1 + 1 ) = ( a + b ) 2 + ( b − a ) 2
Now, we let ( a + b ) 2 + ( b − a ) 2 = 1 6 8 8 4 2 to get
2 ( a 2 + b 2 ) = ( a + b ) 2 + ( b − a ) 2 = 1 6 8 8 4 2 or a 2 + b 2 = 8 4 4 2 1
Now, from the format above, we see that a + b = 3 1 9 , b − a = 2 5 9
So, a + b = 3 1 9
very nice question..
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Note that 8 4 4 2 1 = 2 1 ⋅ 1 6 8 8 4 2 ; a 2 + b 2 = 2 1 ( 2 5 9 2 + 3 1 9 2 ) ; a 2 + b 2 = 2 1 ( ( a − b ) 2 + ( a + b ) 2 ) .
Thus we set a − b = 2 5 9 , a + b = 3 1 9 . It is easy to check that a = 2 8 9 and b = 3 0 are indeed integers.