Cubic with integer solutions
Let f ( x ) = x 3 − a x 2 + b x − b for some positive integers a and b . If the roots of f ( x ) = 0 are distinct positive integers, what is the value of a + b ?
The answer is 47.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
How do you find the three solutions?I just noticed that (2,3,6) is one so I got the right answer.
Log in to reply
WLOG, let r 1 ≤ r 2 ≤ r 3 . Notice that r 1 cannot be greater than 3 , or else the greatest possible sum is 4 1 + 4 1 + 4 1 = 4 3 < 1 . Obviously, r 1 = 1 .
If r 1 = 3 , our only solution is ( 3 , 3 , 3 ) .
If r 1 = 2 , we have
r 2 1 + r 3 1 r 2 r 3 r 2 + r 3 2 ( r 2 + r 3 ) r 2 r 3 − 2 r 2 − 2 r 3 ( r 2 − 2 ) ( r 3 − 2 ) = 2 1 = 2 1 = r 2 r 3 = 0 = 4 .
Since r 2 and r 3 are integers, we get r 2 = 4 , r 3 = 4 and r 2 = 3 , r 3 = 6 . So, we get the solutions ( 2 , 4 , 4 ) and ( 2 , 3 , 6 ) .
Log in to reply
I see, thanks for making that clear.
Nice explaination , I did in brute force method . :)
Did the exact same :)
Really nice application of Vieta's formulas.
Let the roots be r 1 , r 2 , and r 3 . From Vieta's, we get r 1 r 2 + r 2 r 3 + r 3 r 1 = b and r 1 r 2 r 3 = b . Dividing the first equation by the second yields
r 1 1 + r 2 1 + r 3 1 = 1 .
If we solve this equation for positive integers, our only solutions turn out to be ( 3 , 3 , 3 ) , ( 2 , 4 , 4 ) , ( 2 , 3 , 6 ) , and permutations of these. Since r 1 = r 2 = r 3 , our polynomial must have roots 2 , 3 , and 6 . This immediately gives a = 2 + 3 + 6 = 1 1 . To find b we use Vieta's to get b = r 1 r 2 r 3 = 2 ( 3 ) ( 6 ) = 3 6 . Thus, a + b = 1 1 + 3 6 = 4 7 .