Diophantine with single-digit unknowns

Number Theory Level 4

Given that Diophantine equation a x 2 + b x = ( a + b + c ) 2 + a + b \large ax^2+bx=(a+b+c)^2+a+b where a a , b b , and c c are different single-digit natural numbers (i.e. 1 to 9) has one root equal (of x x ) to 10, what is the other root?


The answer is -15.

Rajen Kapur by Rajen Kapur , 72, India

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1 solution

Rajen Kapur
Feb 13, 2017

Put x = 10 x=10 100 a + 10 b = ( a + b + c ) 2 + a + b → 9 ( 11 a + b ) = ( a + b + c ) 2 100a+10b=(a+b+c)^2+a+b \rightarrow 9(11a+b)=(a+b+c)^2 Let a + b + c = 3 t a+b+c=3t 11 a + b = t 2 → 10 a − c = t ( t − 3 ) 11a+b=t^2 \rightarrow10a-c=t(t-3) Unique solution in single-digits is t = 4 t=4 , a = 1 a=1 , c = 6 c=6 , and consequently b = 5 b=5 . Now putting back these values in the equation, we get x 2 + 5 x = 150 → ( x − 10 ) ( x + 15 ) = 0 x^2+5x=150 \rightarrow(x-10)(x+15)=0 Hence the other root is -15 \fbox {-15} .

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