The graphs $y=3(x−h)^2+j$ and $y=2(x−h)^2+k$ have y-intercepts of $2013$ and $2014$, respectively, and each graph has two positive integer x-intercepts. Find $h$.
The answer is an integer between 1 and 999.
I substituted $0$ for $x$ in both equations and I was able to derive that $2016=3k-2j.$ I am not really sure where to go from here.
On
Here is a hint which should let you finish off the problem: Note that each parabola has positive integer x-intercepts. So for some positive integer n, $$3(n-h)^2+j=0\implies j = -3(n-h)^2$$ So j is a square multiplied by -3. Similarly, k is a square multiplied by -2. So let $$j=-3a^2, k=-2b^2$$ Now $$3k-2j = 2016 \implies 6a^2 - 6b^2 = 2016 \implies (a+b)(a-b) = 336$$ You should now consider factors of 336 to find some values a,b and then test which ones work using $$3h^2 + j = 2013, 2h^2 + k = 2014$$ Hope this helps.
We are given that there are two positive integers $x$ such that $3(x-h)^2 + j =0$. For one of those values of $x$, let $p = x-h$; then we must have $j = -3p^2$.
Similarly, we must have $k = -2q^2$ for some integer $q$.
Now, when we set $x=0$ to find the $y$-intercept, we have $$3(0-h)^2 - 3p^2 = 2013, \qquad 2(0-h)^2 - 2q^2 = 2014$$ or $h^2-p^2 = 671$ and $h^2-q^2 = 1007$.
These both factor as differences of squares: $(h+p)(h-p)=671$ and $(h+q)(h-q)=1007$. Coincidentally, both $671$ and $1007$ have very few factorizations: $671 = 671 \cdot 1 = 61 \cdot 11$ and $1007 = 1007 \cdot 1 = 53 \cdot 19$.
Setting $h+p = 61$, $h-p = 11$, $h+q = 53$, $h-q = 19$ turns out to be the only alternative which works. It gives us $h = 36$, $p = 25$ (which means $j=-1875$), and $q=17$ (which means $k = -578$).