Sequences & Series
Integer pairs with AM = GM + 8
MJAT_TS3_P1
Grade 12
Question:
Let $a,b$ be two integers such that $0<a<b<10^6$ and the arithmetic mean of $a$ and $b$ is exactly $8$ more than its geometric mean. If the number of such ordered pairs is $N$, then $N$ equals:
Step-by-Step Solution
Key Concept: $\frac{a+b}{2} = \sqrt{ab}+8$. Let $\sqrt{a}=m$, $\sqrt{b}=n$ with $m<n$ (both positive integers since $a,b$ must be perfect squares for AM-GM to be exactly 8 more). Then $m^2+n^2=2mn+16\Rightarrow(n-m)^2=16\Rightarrow n-m=4$.
$N=\mathbf{995}$.
Correct Answer: 995