SOLUTION: Prove that there exists some integer n such that n^2 + 123457 is a perfect square.

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Question 1179155: Prove that there exists some integer n such that n^2 + 123457 is a perfect square.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose there is integer m so that

n%5E2%2B123457=m%5E2

123457=m%5E2-n%5E2

123457=%28m-n%29%28m%2Bn%29

Then since 123457=(1)(123457),

Those factors m-n and m+n could be 1 and 123457 respectively.

We have the system

system%28m-n=1%2Cm%2Bn=123457%29

m = 1+n

Substitute in

    m+n = 123457
1+n + n = 123457
   2n+1 = 123457
     2n = 123456
      n = 61728

m = 1+n = 61729

So n2 + 123457 = 617282 + 123457 = 

617282 + 123457 = 627192, which is a perfect square.

Edwin