+0

# result on polynomials

-1
5
2

Let $f(x)$ be a polynomial with integer coefficients. There exist distinct integers $p,$ $q,$ $r,$ $s,$ $t$ such that
$f(p) = f(q) = f(r) = f(s) = 1$
and $f(t) > 1.$ What is the smallest possible value of $f(t)?$

Aug 24, 2023

#2
+1

I apologize, but I have to amend my answer because I made an oversight that I cannot leave uncorrected.To minimize $$(t - p)(t - q)(t - r)(t - s)$$ where $$p, q, r, \text{ and } s$$ are all distinct integers, I previously suggested $$1 * 2 * 3 * 4$$ as a product of four integers that minimizes the product but ensuring the result is positive, but I must not forget about the negative integers! With the negative numbers, the minimized product would be $$1 * -1 * 2 * -2$$. Notice that since 2 of the integers are negative that the end result will indeed be positive. With this correction, $$g(x) = (x - 1)(x + 1)(x - 2)(x + 2)$$, where t is minimized when t = 0, so $$g(t) = g(0) = -1 * 1 * -2 * 2 = 4$$, which is significantly less than 24 from my previous answer!

Now, perform the transformation as I described in my previous answer. That would result in $$f(x) = (x - 1)(x + 1)(x - 2)(x + 2) + 1$$, so $$f(t) = f(0) = -1 * 1 * -2 * 2 + 1 = 5$$.

Aug 26, 2023

#1
0

One strategy is to think of the "easiest" version of the problem and then applying the appropriate modifications to the true question at hand. I imagined a similar polynomial function $$g(x)$$ with the same constraints except that $$g(p) = g(q) = g(r) = g(s) = 0$$. I made this slight adjustment so that I can take advantage of the fact that now $$p, q, r, \text{ and } s$$ are all now roots of $$g(x)$$. Because of this, we can write $$g(x) = a(x - p)(x - q)(x - r)(x - s)$$. Now, our goal is to minimize but ensure that $$g(t) > 0$$.

As a result, we know that$$g(t) = a(t - p)(t - q)(t - r)(t - s) > 0$$. Note that the original problem states that $$p, q, r, \text{ and } s$$ are all distinct integers, so  $$t - p, t - q, t - r, \text{ and } t - s$$ are all unique intergers involved in a positive product. We want to pick these integers strategically such that the product is minimized but still positive. By observation, I think it is clear that $$1 * 2 * 3 * 4$$ is the smallest product of four unique integers that is also positive. The idea is to pick the smallest integers possible so that the product is the smallest possible. Now, we must pick a value for $$a$$. Once again, we want the product to be as small as possible while still remaining positive and ensuring that the polynomial has integer coefficients, so picking $$a = 1$$ is the best choice.

We have fantasized about a mystery polynomial that has these properties, but can we make this dream a reality? Yes! All we have to do is craft a polynomial with these properties. There are infinitely many that would satisfy our dreams, but one of the simplest is to write $$g(x) = (x + 1)(x + 2)(x + 3)(x + 4)$$. Now, $$g(0) = 1 * 2 * 3 * 4 = 24$$. In this example, $$a = 1, p = -1, q = -2, r = -3, s = -4, \text{ and } t = 0$$. This fits all the criteria and minimizes $$g(0)$$.

While we were successful in generating $$g(x)$$, we still have to apply this problem to $$f(x)$$. Luckily, this is not a problem because we can take advantage of simple transformations to make the characteristics of $$g(x)$$ apply to $$f(x)$$. We can observe that if $$f(x) = g(x) + 1$$, then everything still holds. This would make$$f(x) = (x + 1)(x + 2)(x + 3)(x + 4) + 1$$, and $$f(t)$$ is still minimized at $$t = 0$$. Therefore, $$f(t) = f(0) = (0 + 1)(0 + 2)(0 + 3)(0 + 4) + 1= 1 * 2 * 3 * 4 + 1 = 25$$.

We are done!

Aug 25, 2023
#2
+1
I apologize, but I have to amend my answer because I made an oversight that I cannot leave uncorrected.To minimize $$(t - p)(t - q)(t - r)(t - s)$$ where $$p, q, r, \text{ and } s$$ are all distinct integers, I previously suggested $$1 * 2 * 3 * 4$$ as a product of four integers that minimizes the product but ensuring the result is positive, but I must not forget about the negative integers! With the negative numbers, the minimized product would be $$1 * -1 * 2 * -2$$. Notice that since 2 of the integers are negative that the end result will indeed be positive. With this correction, $$g(x) = (x - 1)(x + 1)(x - 2)(x + 2)$$, where t is minimized when t = 0, so $$g(t) = g(0) = -1 * 1 * -2 * 2 = 4$$, which is significantly less than 24 from my previous answer!
Now, perform the transformation as I described in my previous answer. That would result in $$f(x) = (x - 1)(x + 1)(x - 2)(x + 2) + 1$$, so $$f(t) = f(0) = -1 * 1 * -2 * 2 + 1 = 5$$.