Gil is 17 years older than Sheila. If his age is written after hers, the result is a 4-digit perfect square. The same statement will also be true 13 years from now. Find Sheila’s present age.
The problem suggests a whole lot of guess-and-check, but let’s see if we can use some algebra.
Let s represent Sheila’s age and s + 17, Gil’s age.
We observe that 100s + s + 17 makes the 4-digit perfect square, say k^{2}.
100s + s + 17 = k^{2} → 101s + 17 = k^{2}.
Now, in 13 years we’ll have 100(s + 13) + (s + 17 + 13) = m^{2} for some m.
100s + 1300 + s + 30 = m^{2} → 101s + 1330 = m^{2}.
So: | 101s + 1330 | = | m^{2} | |
101s + 17 | = | k^{2} | ||
1313 | = | m^{2} – k^{2} | = (m + k)(m – k) |
1313 = 101 · 13, both factors prime.
So 101 · 13 = (m + k)(m – k) → | m + k | = | 101 | |
m – k | = | 13 | ||
2m | = | 114 | → m = 57 and k = 44. |
We’re done!
44^{2} = 1936: s = 19 and Gil is 36. (19 + 17 = 36)
And in 13 years, Sheila is 32 and Gil is 49, and 3249 = 57^{2}.
And in 13 years, Sheila is 32 and Gil is 49, and 3249 = 57^{2}.