A problem by OJAS JAIN

Level pending

How many integer pairs (x,y) satisfy x 2 + 4 y 2 2 x y 2 x 4 y 8 = 0 x^2+4y^2-2xy-2x-4y-8=0


The answer is 6.

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1 solution

Jack D'Aurizio
Apr 19, 2014

By setting A = x y , B = y A=x-y,B=y we can put the original equation in the form ( A 1 ) 2 + 3 ( B 1 ) 2 = 12 (A-1)^2+3(B-1)^2=12 . Since the only integer solutions of C 2 + 3 D 2 = 12 C^2+3D^2=12 are ( C , D ) { ( 0 , ± 2 ) , ( ± 3 , ± 1 ) } (C,D)\in\{(0,\pm 2),(\pm 3,\pm 1)\} , just six pairs of integers satisfy the original equation: ( 2 , 0 ) , ( 0 , 1 ) , ( 0 , 2 ) , ( 4 , 0 ) , ( 4 , 3 ) , ( 6 , 2 ) (-2,0),(0,-1),(0,2),(4,0),(4,3),(6,2) .

ITS NICE BUT HOW YOU GOT THAT ... A =X-Y AND B=Y-1??????????????????? HOW THZ THING CAME IN YOUR HEAD/ PLZ TELLL........

Rohit Singh - 7 years, 1 month ago

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