a^2 + 48 = b^2
b^2 + 48 = c^2
since (n+1)^2 = n^2+2n+1, 2n+1 must be less than 48, so n<24
If a is odd, it must end in 1 or 9 for a^2+48 to be a square. If a or b is greater than 24, the next possible greater square is more than 48 away.
1^2+48 = 7^2
11^2+48 = 13^2
but these don't end in 1 or 9, so adding 48 cannot produce a square
so, a is even and must end in 6
Hmmm. No good.
Sorry. I don't see a solution.
The population in a town called Atown in December 1900 is a perfect square. In January 1901, 48 people moved in, and the population is 1 more than a perfect square. In February 1901, 48 more people moved in, and the population is a perfect square. Assuming that no one was born, died or moved out during these months, what is the population of Atown in December 1900?
3 answers
Got it!
Initial population is 23^2, which = 529.
23^2 + 48 = 577 = 576 + 1, so
23^2 + 48 = 24^2 + 1, and
23^2 + 96 = 625, so
23^2 = 25^2
Did it by exhaustion with Excel. Check back later if I can do it algebraicly!
Initial population is 23^2, which = 529.
23^2 + 48 = 577 = 576 + 1, so
23^2 + 48 = 24^2 + 1, and
23^2 + 96 = 625, so
23^2 = 25^2
Did it by exhaustion with Excel. Check back later if I can do it algebraicly!
I mean 23^2 + 96 = 25^2
2 answers