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
The formula for finding the sum of all positive integers a is (n(n+1))/2, where n is the highest positive integer in the series.
To find the sum of all positive integers a using a mathematical series, use the formula Sn = (n(a1 + an))/2, where n is the number of terms in the series, a1 is the first term, and an is the last term.
Yes, you can use a calculator to find the sum of all positive integers a by plugging in the values into the formula (n(n+1))/2 or Sn = (n(a1 + an))/2.
Yes, there is a shortcut or trick to finding the sum of all positive integers a. You can use the formula (n(n+1))/2 to find the sum without having to manually add all the numbers in the series.
No, it is not possible to find the sum of all positive integers a if the series is infinite. The sum would approach infinity and cannot be calculated accurately.