Deadly Problem
Given that, x 4 y 3 + x 3 y 2 + x 2 y + x =139655 such that x , y are Natural Numbers. Find the minimum values of x , y which satisfy above condition , if, 3 x + 2 y = 2 7 . Then find x + y .
The answer is 11.
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1 solution
Very correct approach , but can we do it by any identity? I did it in same way as you did.
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Noting that f ( x , y ) = x 4 y 3 + x 3 y 2 + x 2 y + x is a geometric progression we can factor it to
f ( x , y ) = x y − 1 x ( ( x y ) 4 − 1 ) = x ( ( x y ) 2 + 1 ) ( x y + 1 ) meaning 2 9 1 8 7 1 5 5 is divisible by x
Noting that because 3 x + 2 y = 2 7 we have x < 1 0 , simple divisibility checks show 2 9 1 8 7 1 5 5 is only divisible 1 and 5 under 1 0 . Trying 1 , f ( 1 , y ) = ( y 2 + 1 ) ( y + 1 ) . Even for y = 1 0 0 this value has only 7 digits so x = 5 , consequently y = 6 and the answer is 1 1
Another more, formulaic and identity based method, with less case work if thats what you were looking for!
The natural number solutions of the equation 3 x + 2 y = 2 7 must be one of the five options ( x , y ) = ( 9 , 0 ) , ( 7 , 3 ) , ( 5 , 6 ) , ( 3 , 9 ) , ( 1 , 1 2 ) .
Define f ( x , y ) = x 4 y 3 + x 3 y 2 + x 2 y + x
If both x and y are odd, the each term will be odd, as there are four terms, f ( x , y ) would be even, ruling out the second and fourth options. f ( 9 , 0 ) = 0 + 0 + 0 + 9 , rules out the first leaving two possible solutions, ( x , y ) = ( 5 , 6 ) , ( 1 , 1 2 )
f ( 1 , 1 2 ) = 1 2 3 + 1 2 2 + 1 2 < 2 0 3 + 2 0 2 + 2 0 = 8 4 2 0 < 2 9 1 8 7 1 5 5
Therefore ( x , y ) = ( 5 , 6 ) making the answer 5 + 6 = 1 1