The problem requires finding the sum of all positive integers \( a \) such that the expression \( \sqrt{\sqrt{(a+500)^2-250000}-a} \) results in an integer. The key steps involve simplifying the expression and determining the conditions under which the inner square root yields a non-negative integer. The solution reveals specific values of \( a \) that satisfy these conditions, leading to a definitive sum of these integers.
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anemone said:Find the sum of all positive integers $a$ such that $\sqrt{\sqrt{(a+500)^2-250000}-a}$ is an integer.
Albert said:let m=$\sqrt{\sqrt{(a+500)^2-250000}-a}----(*)$
n=$\sqrt{(a+500)^2-250000}< a+500$
$m<\sqrt{500}\,\, or\,\, m\leq 22$---(1)
$n=\sqrt {a(a+1000)}=\sqrt{a(a+2^35^3)}$
min(a)=125=$5^3$
and $\sqrt {250}<m \,\, or \, 16\leq m$----(2)
from (1) and (2) put m=16,17,18,19,20,21,22 to (*)
we get m=20 where $a=800=2^55^2$ is the only solution