Find the least possible value of $k$ for $(2y-2017)^2 = k$ given the constraints.

Question

The equation $(2y-2017)^2=k$, where $k \in \mathbb R$, has two distinct positive integer solutions for $y$, one of which is a multiple of $100$. What is the least possible value of $k$?

Answer 1

We can solve the equation for $y$ to find $$|2y - 2017| = \sqrt{k},$$ and so the two solutions for $y$ are $$2y - 2017 = \sqrt{k}$$ which gives $$y = \frac{2017 + \sqrt{k}}{2},$$ and $$2y - 2017 = -\sqrt{k},$$ which gives $$y = \frac{2017 - \sqrt{k}}{2}.$$

Since these two values are positive integers, we must have that $2017 - \sqrt{k} > 0$ (for the second solution to be positive) and that $\sqrt{k}$ is an odd integer (so that $2017 - \sqrt{k}$ is an even integer, which is required since it divided by $2$ is the integral second solution).

Now, the multiple of $100$ bit.

If the second solution is the multiple of $100,$ then $\sqrt{k} \equiv 17 \pmod{200},$ and so again we must have $k \equiv 89\pmod{200},$ giving us a least solution of $289$.

If the first solution is a multiple of $100,$ then $2017 + \sqrt{k}$ is a multiple of $200$ (since it divided by $2$ is a multiple of $100$). $2017 \equiv 17 \pmod{200},$ and so $\sqrt{k}$ must be $-17$ modulo $200.$ So, $k \equiv (-17)^2 \equiv 89 \pmod{200}.$ $89$ is not a perfect square, $289$'s square root isn't $-17$ modulo $200,$ and the next thing we'd have to check is larger than $289.$ So, the least value of $k$ is $289.$

Answer 2

(Not the best way ) but an elementary approach for you :

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Since $K$ is odd and is a perfect square , we can write $K = (2m-1)^2$ , for some $m\in \mathbb N.$ So , the equation is reduced to :

$$(2y-2017) = \pm (2m-1)$$ giving us ,

$$\color{red}{y-1008 = m} \quad \text{or} \quad \color{blue}{y+m=1009}$$

Since one value of $y$ is a multiple of $100$ , we can try inputting some values of $y$ to determine $m$ .

Taking $y-1008=m$ , we want to find the minimum value of $m$ , so we take $y = 1100$ , which yields : $\quad\color{#2dd}{m = 92 }. $

Taking $y+m = 1009$ and again to minimize $k$ and also to minimize $m$ , we take $y = 1000 $ yielding :$\quad\color{#2d0}{m = 9}$ .

So the smallest value we got is $m = 9.$ Putting this value into $K =(2m-1)^2$ , we have the value of $\boxed{k = 289}$