Question 1179155
<pre>
Suppose there is integer m so that

{{{n^2+123457=m^2}}}

{{{123457=m^2-n^2}}}

{{{123457=(m-n)(m+n)}}}

Then since 123457=(1)(123457),

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

We have the system

{{{system(m-n=1,m+n=123457)}}}

m = 1+n

Substitute in

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

m = 1+n = 61729

So n<sup>2</sup> + 123457 = 61728<sup>2</sup> + 123457 = 

61728<sup>2</sup> + 123457 = 62719<sup>2</sup>, which is a perfect square.

Edwin</pre>