SOLUTION: The sum of the first m positive odd integers is 212 more than the sum of the first n positive even integers. What is the sum of all possible values of n?

Question 1148539: The sum of the first m positive odd integers is 212 more than the sum of the first n positive even integers. What is the sum of all possible values of n?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The sum of the first m positive odd integers is m^2.

The sum of the first n positive even integers is n^2+n.

So the requirement is that m^2 = n^2+n+212; i.e., the square root of (n^2+n+212) must be an integer.

I don't know an algebraic way to find solutions.

An excel spreadsheet for calculating the square root of n^2+n+212 shows three solutions:

n = 16; m = 22
n = 28; m = 32
n = 211; m = 212

So the answer to the given question is 16+28+211 = 255.

I will be interested to see if any of the other tutors have an algebraic method for finding those solutions....
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.

For algebraic solutions to this problem, see this link

https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_22

https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_22

Solution 3.

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