Respuesta :
The difference between consecutive integral squares must be greater than 1000. (x+1)^2-x^2>=1000, so x>=999/2 implies x>=500. x=500 does not work, so x>500. Let n=x-500. The sum of the square of n and a number a little over 1000 must result in a new perfect square. By inspection, n^2 should end in a number close to but less than 1000 such that there exists 1000N within the difference of the two squares. Examine when n^2=1000. Then, n=10sqrt(10). One example way to estimate sqrt(10) follows.
3^2=9, so 10=(x+3)^2=x^2+6x+9. x^2 is small, so 10=6x+9. x=1/6 implies sqrt(10)\approx 19/6. This is 3.16.
Then, n approx 31.6. n^2<1000, so n could be 31. Add 500 to get the first square and 501 to get the second. Then, the two integral squares are 531^2 and 532^2. Checking, 531^2=281961 and 532^2=283024. 282,000 straddles the two squares, which have a difference of 1063. The difference has been minimized, so N is minimized N=282000 implies 282
for more information visit this link
https://brainly.com/question/20899279
#SPJ4
