dx + dy = d(x + y) = 1000 \Rightarrow d \mid 1000

February 22, 2026

$dx + dy = d(x + y) = 1000 \Rightarrow d \mid 1000$

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r = \frac{84}{21} = 4
$$Question: What is the largest possible value of $\gcd(a,b)$, if the sum of two positive integers $a$ and $b$ is $1000$ and both $a$ and $b$ are divisible by a perfect square greater than $1$?
Let $d = \gcd(a,b)$. Then we can write $a = dx$, $b = dy$, where $\gcd(x,y) = 1$. Since $a + b = 1000$, we have:
So $d$ must be a divisor of 1000. Also, both $a$ and $b$ must be divisible by a perfect square greater than 1. Since $a = dx$ and $b = dy$, both $dx$ and $dy$ must be divisible by a square $s > 1$. That means $d$ must contain all the square factors of $s$, or at least $x$ and $y$ must together contain the square factors.
To maximize $d$, we look at the largest divisor of 1000 such that $x = \frac{a}{d}$ and $y = \frac{b}{d}$ are coprime and at least one of $x$ or $y$ is divisible by a square greater than 1.
The prime factorization of 1000 is:

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