MHB Prove that a function has no integer roots

AI Thread Summary
The function f(x) = x^4 - px^3 - qx^2 - rx - s, where p, q, r, s are natural numbers with p ≥ q ≥ r ≥ s, has no integer roots. For positive integers, assuming f(n) = 0 leads to a contradiction as the left side exceeds the right side when evaluated. For negative integers, the analysis shows that the left side remains greater than the right side, resulting in another contradiction. Thus, both cases confirm that there are no integer solutions for the function. The conclusion is that f(x) cannot have any integer roots.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Let $p, q, r, s \in \mathbb{N}$ such that $p \ge q \ge r \ge s$. Show that the function $f(x)=x^4-px^3-qx^2-rx-s$ has no integer root.
 
Last edited:
Mathematics news on Phys.org
anemone said:
Let $p, q, r, s \in \mathbb{N}$ such that $p \ge q \ge r \ge s$. Show that the function $f(x)=x^4-px^3-qx^2-rx-s$ has no integer root.
[sp]Suppose that $n$ is an integer such that $f(n) = 0$. Then $n\ne0$, because $s\ne0$.

Suppose that $n>0$. Then the equation can be written $n^4 = pn^3+qn^2+rn+s$. We must have $p<n$, otherwise the right side would be greater than the left. Hence $p,q,r,s$ are all $\leqslant n-1$. Therefore $n^4 \leqslant (n-1)n^3 + (n-1)n^2 + (n-1)n + (n-1) = n^4-1.$ That is a contradiction, so there cannot be any positive solutions.

Suppose that $n<0$. Then the equation can be written $n^4 + p|n|^3 +r|n| = qn^2 + s$. But $n^4>0$, $p|n|\geqslant q$ and $r|n|\geqslant s$. So the left side of the equation is greater than the right, and again we have a contradiction.

Therefore there are no integer solutions.[/sp]
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Replies
2
Views
1K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
7
Views
2K
Back
Top