Finding $k$ such that $x^2 + x + 10 = k(k-1)$ has one positive integer root

Question

Suppose $k$ is a positive integer, such that $x^2 + x + 10 = k(k-1)$ has one positive integer root. Find $k.$

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I've tried to factor this and apply the discriminant, but I'm not sure how to deal with the part about the positive integer root. Can I have a hint please?

Answer 1

Notice that this is a quadratic equation in $x$ with solutions $$\frac12\cdot\left(-1 \pm\sqrt{-39 - 4 k + 4 k^2}\right)$$ For one of the root to be an integer, you need $4k^2-4k-39=m^2\iff (2k-1)^2-40=m^2$ for some $m\in\mathbb N$. Have a look below if this is still insufficient

Observe that the last equation is equivalent to $$(2k-1)^2-m^2=40\iff (2k-1+m)(2k-1-m)=40$$ Consider the factor of 40, and check whether the respective values of $k$ satisfy the problem's statement.

Answer 2

So, lets see what we get with the discriminant condition:

$$ 0 = D$$

$$ 0 = 1- 4(10-k(k-1) )$$

$$ 40- 4k (k-1) =1$$

$$ 39 -4k^2 + 4k = 0$$

Now, need to find set of $k$ satisfying the above relation. It can be found that if $a$ and $b$ are the roots of the quadratic in $k$ then set of $k$ values possible are k $ \in [a,b]$ (*)

In the set of $ [a,b]$ simply write down all the positive integers in it and you are done!

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If

$$ f(k) = 39 - 4k^2 + 4k$$

Then the limit:

$$ \lim_{ k \to \infty} f(k) = -ve$$

It is negative because the $ -4k^2$ term dominates. With that in mind, the sign flip to become positive happens at a root, then it is undone at the next root. So, between the root is region where $k$ is +ve