# AIME Parabola Question

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.

• I'm curious whether this is a typo $\color{blue}{3}(x-h)\color{red}{2} + j$ – caverac May 30 '18 at 23:34
• This red $2$ is certainly an exponent. We have two parabolas. – Piquito May 30 '18 at 23:36
• Probably should be $3(x - h)^2 + j$ – Phil H May 30 '18 at 23:37
• In that case $3k - 2j = 2016$ – Phil H May 30 '18 at 23:42
• oops i'm bad at latex and at adding apparently – Dude156 May 30 '18 at 23:51

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 $k = -578$).

• Thanks for the cool solution! That was pretty smart! – Dude156 May 31 '18 at 0:29

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.

• That was pretty ingenious! Thanks for the help! – Dude156 May 31 '18 at 0:14
• Btw, how did you come to thought of substituting a instead of that n-h stuffs? – Dude156 May 31 '18 at 0:15
• You don't have to substitute a, and you can leave it as n-h. However, for the sake of simplicity, since n is unknown and arbitrary, I thought it was better to substitute it for a (and the other for b). I'm glad that I was able to help! – Ethan May 31 '18 at 0:17