Smallest possible value of $k$ such that the roots of $x^2-127x+k=0$ are positive integers

Question

In a triangle, two sides have equal lengths both shorter than the third side. The length of the three sides are all integers and all satisfy the equation $x^2-127x+k=0$, $k$ is a constant. Find the smallest possible value of $k$.

Discriminant $(Δ)$ of the quadratic equation: $Δ = b^2 - 4ac.$ In this case, $a = 1$, $b = -127$, and $c = k$, so $Δ = (-127)^2 - 4(1)(k) = 16129 - 4k.$

$Δ > 0 \implies 16129 - 4k > 0.$

Solving for $k$: $4k < 16129 \implies k < 4032.25.$

Since $k$ must be an integer, which is $4031$.

Where did I get wrong?

Answer 1

You already have the reason why your approach isn't enough from the comments. Below is an approach that will find the smallest $k$.

Let $a$ be the length of the short sides and $a+x$ be the length of the longer one. Then $0<x<a$ for a non-degenerate triangle.

From Vieta's relations, we have $2a+x=127$, and $k=a(a+x)=\frac14(127^2-x^2)$. Thus $k$ will be minimised when we maximise $x$. Putting $x=a-1$, we get $3a-1=127\implies a\not \in \mathbb Z$. Next we have $x=a-2 \implies 3a-2=127$ giving $(a, b) = (43, 84)$ and $k_{min}=3612$.