Let a , b be coprime positive integers are such that none of a , b and a + b are divisible by 7 , but ( a + b ) 7 − a 7 − b 7 is divisible by 7 7 . Find the minimum value of a + b .
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Hi Chew-Seong, you're making a false assumption here. We CANNOT assume that a 2 + b 2 + a b must be equal to 7 3 . All we can gather is that a 2 + b 2 + a b is divisible by 7 3 .
In fact, there are infinitely many solutions of ( a , b ) . Here are some of the solutions: ( a , b ) = ( 1 , 1 8 ) , ( 1 , 3 2 4 ) , ( 1 , 3 6 1 ) , ( 2 , 3 0 5 ) , ( 2 , 3 7 9 ) , ( 3 , 2 8 6 ) , ( 3 , 3 9 7 ) , ( 4 , 2 6 7 ) , ( 5 , 2 4 8 ) , ( 6 , 2 2 9 ) , ( 8 , 1 9 1 ) , ( 9 , 1 7 2 ) , ( 1 0 , 1 5 3 ) .
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( a + b ) 7 − a 7 − b 7 7 a 6 b + 2 1 a 5 b 2 + 3 5 a 4 b 3 + 3 5 a 3 b 4 + 2 1 a 2 b 5 + 7 a b 6 a 6 b + 3 a 5 b 2 + 5 a 4 b 3 + 5 a 3 b 4 + 3 a 2 b 5 + a b 6 a 5 + 3 a 4 b + 5 a 3 b 2 + 5 a 2 b 3 + 3 a b 4 + b 5 a 5 + b 5 + 3 a b ( a 3 + b 3 ) + 5 a 2 b 2 ( a + b ) ( a + b ) ( a 4 − a 3 b + a 2 b 2 − a b 3 + b 4 ) + 3 a b ( a + b ) ( a 2 − a b + b 2 ) + 5 a 2 b 2 ( a + b ) a 4 − a 3 b + a 2 b 2 − a b 3 + b 4 + 3 a b ( a 2 − a b + b 2 ) + 5 a 2 b 2 a 4 + b 4 + 2 a b ( a 2 + b 2 ) + 3 a 2 b 2 ( a 2 + b 2 ) 2 + 2 a b ( a 2 + b 2 ) + a 2 b 2 ( a 2 + b 2 + a b ) 2 a 2 + b 2 + a b ≡ 0 (mod 7 7 ) ≡ 0 (mod 7 7 ) ≡ 0 (mod 7 6 ) ≡ 0 (mod 7 6 ) ≡ 0 (mod 7 6 ) ≡ 0 (mod 7 6 ) ≡ 0 (mod 7 6 ) ≡ 0 (mod 7 6 ) ≡ 0 (mod 7 6 ) ≡ 0 (mod 7 6 ) ≡ 0 (mod 7 3 ) Since a b is indivisible by 7 Since a + b is indivisible by 7
Assuming that
a 2 + b 2 + a b ⟹ b 2 + a b + a 2 − 3 4 3 ⟹ b = 7 3 = 3 4 3 = 0 = 2 − a + 1 3 7 2 − 3 a 2
For b to be an integer, 1 3 7 2 − 3 a 2 must be a perfect square. Solving for a , we have the unordered pairs of ( a , b ) = ( − 7 , 2 1 ) , ( − 1 , 1 9 ) , ( 1 , 1 8 ) , ( 7 , 1 4 ) . Therefore there is only one acceptable solution and a + b = 1 + 1 8 = 1 9 .