More sums of squares
5 4 8 4 1 = 3 1 7 × 1 7 3 . 54841, whose prime factor decomposition is given above, can be written as the sum of two perfect squares in two different ways:
5 4 8 4 1 = a 2 + b 2 = c 2 + d 2 ,
where a , b , c and d are distinct positive integers. Find the sum a + b + c + d .
Hint : 3 1 7 = 1 9 6 + 1 2 1 , and 1 7 3 = 1 6 9 + 4 .
The answer is 650.
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2 solutions
Working in the Gaussian integers we have 5 4 8 4 1 = 3 1 7 × 1 7 3 = ( 1 9 6 + 1 2 1 ) ( 1 6 9 + 4 ) = ( 1 4 + 1 1 i ) ( 1 4 − 1 1 i ) ( 1 3 + 2 i ) ( 1 3 − 2 i ) Since these last four numbers are prime in the Gaussian integers, if we want to write 5 4 8 4 1 = X 2 + Y 2 = ( X + i Y ) ( X − i Y ) then (to within association), the possible values of X + i Y are ( 1 4 + 1 1 i ) ( 1 3 + 2 i ) = 1 6 0 + 1 7 1 i ( 1 4 + 1 1 i ) ( 1 3 − 2 i ) = 2 0 4 + 1 1 5 i Thus we have 5 4 8 4 1 = 1 6 0 2 + 1 7 1 2 = 2 0 4 2 + 1 1 5 2 , making the answer 6 5 0 .
Nice use of Gaussian Algebra. It implies that if one sees something of the form ( X 2 + Y 2 ) one should be tempted to think of the form from which it could come which is ( X + i Y ) ( X − i Y ) .
I did it similarly, except I used the notation 5 4 8 4 1 = ∣ ( 1 4 ± 1 1 i ) ( 1 3 + 2 i ) ∣ 2 .
Note that ( u 2 + v 2 ) ( x 2 + y 2 ) = ( u x ± v y ) 2 + ( v x ∓ u y ) 2 . If the two factors on the left are prime numbers (necessarily of the form 4 n + 1 ) then this is a unique decomposition.
Here u = 1 4 , v = 1 1 , x = 1 3 , y = 2 . (This assignment gives positive values.)
Thus a + b + c + d = ( u x + v y ) + ( v x − u y ) + ( u x − v y ) + ( v x − u y ) = 2 ( u + v ) x = 2 ⋅ ( 1 4 + 1 1 ) ⋅ 1 3 = 6 5 0 .