169-follower problem
169 is a square and it is a sum of squares ( 1 6 9 = 1 2 2 + 5 2 ).
But its square can also be written as the sum of two squares. Let 1 6 9 2 = a 2 + b 2 with a > b > 0 and g cd ( a , b ) = 1 . Without using a calculator, find a + b .
The answer is 239.
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4 solutions
By the Fibonacci-Brahmagupta identity,
( a 2 + b 2 ) ( c 2 + d 2 ) = ( a c ± b d ) 2 + ( a d ∓ b c ) 2
So, 1 6 9 2 = ( 1 2 2 + 5 2 ) ( 1 2 2 + 5 2 ) = ( ( 1 2 ) ( 1 2 ) − ( 5 ) ( 5 ) ) 2 + ( ( 1 2 ) ( 5 ) + ( 5 ) ( 1 2 ) ) 2
1 6 9 2 = ( 1 1 9 ) 2 + ( 1 2 0 ) 2
Since g c d ( 1 1 9 , 1 2 0 ) = 1 , therefore
a = 1 1 9 , b = 1 2 0 , a + b = 2 3 9
You need to check that there is no other solution. (In particular, note that you might have possibly picked ( 1 2 ⋅ 1 2 + 5 ⋅ 5 ) 2 + ( 1 2 ⋅ 5 − 5 ⋅ 1 2 ) 2 instead, or you might have used 1 6 9 = 1 3 2 + 0 2 instead of 1 6 9 = 1 2 2 + 5 2 ; you should rule those out as well.
I learnt a lot.Thank You.
We have, 1 6 9 = 1 2 2 + 5 2 Squaring both side yields 1 6 9 2 = ( 1 2 2 + 5 2 ) 2 Or, 1 6 9 2 = ( 1 2 2 − 5 2 ) 2 + 4 ∗ 1 2 2 ∗ 5 2 [Because, ( a + b ) 2 = ( a − b ) 2 + 4 a b ]
Or, 1 6 9 2 = 1 1 9 2 + 1 2 0 2 Since, 4 ∗ 1 2 2 ∗ 5 2 = ( 2 ∗ 1 2 ∗ 5 ) 2 = 1 2 0 2 [P.S. At first, I deduced 1 6 9 2 = 1 5 6 2 + 6 5 2 which doesn't form a primitive pythagorean triple.]
https://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%3D28561
Euclid's formula is a fundamental formula for generating Pythagorean triples given an arbitrary pair of positive integers m and n with m > n. The formula states that the integers a = m 2 − n 2 , b = 2 m n , c = m 2 + n 2 form a Pythagorean triple.
In our case:
c = 1 6 9 = 1 2 2 + 5 2 ⇒ m = 1 2 , n = 5 , a = 1 1 9 , b = 1 2 0 , a + b = 2 3 9