Use your results to find the squares that can be added to 225 to produce another square.
I started off by taking the 9 divides 225 with quotient 25.
(25-8) + (25-6) + (25-4) + (25-2) + 25 + (25+2) + (25+4) + (25+6) + (25+8) = 225
simplifying:
17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 = 225.
Since integers can be negative, some ways will start with negative integers. For example 45 consecutive odd integers can add to 225 like this: (-39) + (-37) + (-35) + … + 45 + 47 + 49 = 225.
So, to answer your question you have to count the number positive odd divisors of 225, and that will be the cardinality of the the set {1, 3, 5, 9, 15, 25, 45, 75, 225}, so the answer is 9.
I'm having trouble with finding the ways to express the sum of consecutive odd integers. I think I am off to a good start but really need a formal answer as well as a way to produce another square. Please help! Thanks!
Best Answer
We know that the sum of the odd numbers up to $2n-1$ is $n^2$. The sum of the odd numbers from $2m+1$ through $2n-1$ is then $n^2-m^2=(n+m)(n-m)$ We need this to be $225$, so you are looking for the number of solutions to $225=(n+m)(n-m)$ Each factorization of $225$ gives you one except $225=15\cdot 15$ (Why?) For example $225=45\cdot 5$ We can then write $n+m=45,n-m=5, n=25, m=20$ and find $225=41+43+45+47+49$. Note that there are $5$ terms in the expression and the middle one is $45$