Algebra Level 3

Let a , b a,b be coprime positive integers are such that none of a a , b b and a + b a+b are divisible by 7 7 , but ( a + b ) 7 a 7 b 7 (a+b)^7-a^7-b^7 is divisible by 7 7 7^7 . Find the minimum value of a + b a+b .


The answer is 19.

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1 solution

( a + b ) 7 a 7 b 7 0 (mod 7 7 ) 7 a 6 b + 21 a 5 b 2 + 35 a 4 b 3 + 35 a 3 b 4 + 21 a 2 b 5 + 7 a b 6 0 (mod 7 7 ) 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 0 (mod 7 6 ) Since a b is indivisible by 7 a 5 + 3 a 4 b + 5 a 3 b 2 + 5 a 2 b 3 + 3 a b 4 + b 5 0 (mod 7 6 ) a 5 + b 5 + 3 a b ( a 3 + b 3 ) + 5 a 2 b 2 ( a + b ) 0 (mod 7 6 ) ( 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 ) 0 (mod 7 6 ) Since a + b is indivisible by 7 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 0 (mod 7 6 ) a 4 + b 4 + 2 a b ( a 2 + b 2 ) + 3 a 2 b 2 0 (mod 7 6 ) ( a 2 + b 2 ) 2 + 2 a b ( a 2 + b 2 ) + a 2 b 2 0 (mod 7 6 ) ( a 2 + b 2 + a b ) 2 0 (mod 7 6 ) a 2 + b 2 + a b 0 (mod 7 3 ) \begin{aligned} (a+b)^7 - a^7 - b^7 & \equiv 0 \text{ (mod 7}^7) \\ 7a^6b + 21a^5b^2 + 35a^4b^3 + 35a^3b^4 + 21a^2b^5 + 7ab^6 & \equiv 0 \text{ (mod 7}^\blue 7) \\ a^6b + 3a^5b^2 + 5a^4b^3 + 5a^3b^4 + 3a^2b^5 + ab^6 & \equiv 0 \text{ (mod 7}^\red 6) & \small \blue{\text{Since }ab\text{ is indivisible by 7}} \\ a^5 + 3a^4b + 5a^3b^2 + 5a^2b^3 + 3ab^4 + b^5 & \equiv 0 \text{ (mod 7}^6) \\ a^5 + b^5 + 3ab(a^3+b^3) + 5a^2b^2(a+b) & \equiv 0 \text{ (mod 7}^6) \\ (a + b)(a^4-a^3b+a^2b^2-ab^3+b^4) + 3ab(a+b)(a^2-ab+b^2) + 5a^2b^2(a+b) & \equiv 0 \text{ (mod 7}^6) & \small \blue{\text{Since }a+b\text{ is indivisible by 7}} \\ a^4-a^3b+a^2b^2-ab^3+b^4 + 3ab(a^2-ab+b^2) + 5a^2b^2 & \equiv 0 \text{ (mod 7}^6) \\ a^4 +b^4 + 2ab(a^2+b^2) + 3a^2b^2 & \equiv 0 \text{ (mod 7}^6) \\ (a^2 +b^2)^2 + 2ab(a^2+b^2) + a^2b^2 & \equiv 0 \text{ (mod 7}^6) \\ \left(a^2 +b^2 + ab\right)^2 & \equiv 0 \text{ (mod 7}^\blue 6) \\ a^2 +b^2 + ab & \equiv 0 \text{ (mod 7}^\red 3) \end{aligned}

Assuming that

a 2 + b 2 + a b = 7 3 = 343 b 2 + a b + a 2 343 = 0 b = a + 1372 3 a 2 2 \begin{aligned} a^2 +b^2 + ab & = 7^3 = 343 \\ \implies b^2 + ab + a^2 - 343 & = 0 \\ \implies b & = \frac {-a + \sqrt{1372-3a^2}}2 \end{aligned}

For b b to be an integer, 1372 3 a 2 1372-3a^2 must be a perfect square. Solving for a a , we have the unordered pairs of ( a , b ) = ( 7 , 21 ) , ( 1 , 19 ) , ( 1 , 18 ) , ( 7 , 14 ) (a,b) = (-7, 21), (-1, 19), \boxed{(1,18)}, (7,14) . Therefore there is only one acceptable solution and a + b = 1 + 18 = 19 a+b = 1 + 18 = \boxed{19} .

Hi Chew-Seong, you're making a false assumption here. We CANNOT assume that a 2 + b 2 + a b a^2 + b^2 + ab must be equal to 7 3 7^3 . All we can gather is that a 2 + b 2 + a b a^2 + b^2 + ab is divisible by 7 3 7^3 .

In fact, there are infinitely many solutions of ( a , b ) (a,b) . Here are some of the solutions: ( a , b ) = ( 1 , 18 ) , ( 1 , 324 ) , ( 1 , 361 ) , ( 2 , 305 ) , ( 2 , 379 ) , ( 3 , 286 ) , ( 3 , 397 ) , ( 4 , 267 ) , ( 5 , 248 ) , ( 6 , 229 ) , ( 8 , 191 ) , ( 9 , 172 ) , ( 10 , 153 ) . (a,b) =(1,18), (1,324), (1,361), (2,305), (2,379), (3,286), (3,397), (4,267), (5,248), (6,229), (8,191), (9,172), (10,153) .

Brilliant Mathematics Staff - 8 months, 2 weeks ago

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