Integer polynomial
Consider all polynomials with integral coefficients
a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 = 0 ,
such that 3 4 is one of its roots, 3 ∣ a 0 , and 4 ∣ a n .
Over all such polynomials, find the smallest positive value of a n + a 0 .
The answer is 12.
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1 solution
Good question @Siddharth Singh I didn't know about that first statement. so I looked it up. It's called Rational root theorem. :)
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Thanks,friend.This theorem was not known to me before,as I came to know about it through a book and simply made a question out of it........Great! never talked IMO and NSO topper before.
@Siddharth Singh ;But if we take a(subscript)0 and a(subscript)n both to be -12,then the condition will be maintained giving the final answer to be -12+(-12)=-24.??
@Siddharth Singh what do you mean by the condition given in the question is reverse??? can you explain more explicit ?
It won't be wrong to assume both 'ao' and 'an' to be zero. Since all integers divide 0, and nothing is mentioned about degree of the Polynomial. Luckily you asked smallest POSITIVE value and that's where 0 was ruled out.
What do you mean by your saying that the condition is reverse? I don't understand, please consider explaining! Thanks!
I was foolish to think you had made a typo in the question. Very good problem. But what would be an example of 'a0' and 'an' that adds up to twelve and satisfies the conditions?
Could you explain why if 12|a0 and 12|an , 12 | (a0 + an)? Wouldn't a0 and an both have to be at least 12 so the smallest value for a0+an would be 24??
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if 1 2 ∣ a 0 ⇒ a 0 = 1 2 q for some integer q and if 1 2 ∣ a n ⇒ a n = 1 2 k for some integer k ; combining the two, we can say a 0 + a n = 1 2 q + 1 2 k = 1 2 ( q + k ) and since q , k are integers, this tells us 1 2 ∣ a 0 + a n ; now, the smallest positive value for this expression is simply the smallest common positive multiple of a 0 , a n , which is clearly 12; hope this helps :)
This can be proved that if a rational number of the form q p is a root of any n-degree equation,then p ∣ a 0 and q ∣ a n .
therefore
4 ∣ a 0 and 3 ∣ a n
but the condition given in the question is reverse and hence the minimum value of both a 0 and a n will be the L.C.M of 4 and 3=12
Hence 1 2 ∣ a 0 and 1 2 ∣ a n ,thus 1 2 ∣ ( a n + a 0 )
the minimum value of a n + a 0 ≥ 1 2