When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square? Pick one:
A) 1296  B) 1369  C) 1681  D) 1764  E) 2500
Let’s call the two perfect squares x^2 and y^2, respectively. Then the given information translates as x^2+148=y^2. Subtracting x^2 gives 148=y^2−x^2, a difference of squares. This, in turn, factors as (y+x)(y−x)=148.   The next step is tricky. It begins with factoring 148, which breaks down as 2∗2∗37. Since we’re dealing with perfect squares, x and y are positive integers, and (y+x) and (y−x) must be paired integer factors of 148. The options are 148∗1,74∗2, and 37∗4. But our number properties establish that (y+x) and (y−x) must be either both odd or both even, so only 74∗2 is an actual possibility. And because for any positive integers (y+x)&gt;(y−x), we can conclude that y+x=74 and y–x=2. Solving by elimination, 2y=76, y=38, and x=36.   Finally, we just need to square 36 . But rather than multiplying it out, note that 36^2 ends in 6 – as does only one answer, choice A. This answer must be the one we want.
<result>A</result>
When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?