X, Y, Dead?

Call an ordered pair of positive integers ( x , y ) (x, y) dead if ( x y ) ( x 2 + x y + y 2 ) x y = 61 (x-y)(x^2+xy+y^2)-xy=61 . Find the sum of all x + y x+y over all dead pairs of positive integers.

If you think there are no pairs of dead integers, enter 0. If you think there are infinitely many pairs of dead integers, enter 1000.

Clarifications : If you think there are 2 pairs of dead integers, (1, 2) and (3, 4), your answer would be ( 1 + 2 ) + ( 3 + 4 ) = 10 (1+2)+(3+4)=10 .


The answer is 11.

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2 solutions

The equation is equivalant to x 3 y 3 x y = 61 x^3-y^3-xy=61 . Now we use the well known factorisation x 3 + y 3 + z 3 = ( x + y + z ) ( x 2 + y 2 + z 2 x y x z y z x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-xz-yz . However, to use this factorisation, we must first multiply both sides by 27 27 to obtain ( 3 x ) 3 + ( 3 y ) 3 + ( 1 ) 3 3 ( 3 x ) ( 3 y ) ( 1 ) = 1646 (3x)^3+(-3y)^3+(-1)^3-3(3x)(-3y)(-1)=1646 . Using the factorisation now, we have ( 3 x 3 y 1 ) ( 9 x 2 + 9 y 2 + 9 x y + 3 x 3 y + 1 ) = 1646 = 2 823 (3x-3y-1)(9x^2+9y^2+9xy+3x-3y+1)=1646=2 \cdot 823 . Notice that the second term must be equivalant to 1 ( m o d 3 ) 1 \pmod 3 , and the first equivalant to 2 ( m o d 3 ) 2 \pmod 3 , so the first term is equal to 2 2 and the second equal to 823 823 . Therefore, we have x y = 1 x-y=1 and 9 x 2 + 9 y 2 + 9 x y + 3 x 3 y + 1 = 823 9x^2+9y^2+9xy+3x-3y+1=823 . We can solve this system of equations to obtain that our only solution is (5, 6). Ans 11 \boxed{11}

Can you close the parentheses on the first row?

. . - 3 months, 3 weeks ago
Rajen Kapur
Jul 23, 2014

x^3 - y^3 = 61 + xy has only one solution x = 6, y = 5. ANSWER x + y = 11

Can you please elaborate.... How do you say that only one pair exists?

Abhinav Raichur - 6 years, 10 months ago

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