One year ago, the number of years in Jane's age was a perfect square and one year from now her age will be a perfect cube. How old is Jane?
Thanks!!
2007-05-15 12:49:59 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Let x = Jane's age
Then x-1 = y²
x+1 =z³.
Subtracting,
z³-y² = 2.
or
y²+2 = z³.
It is well-known that the only integer solution
of this equation is y=5, z =3.
This was proposed by Fermat in the 17th century.
Euler gave a faulty proof of it and a full
proof wasn't given till almost 100 years later.
At any rate, x-1 = 25,
so x = 26.
Jane's age is 26.
To bring this down to 6th grade level, assume
Jane is less than 100 years old and try to find
all other solutions. I don't think many 6th graders
would have the background to prove Fermat's
proposal!
2007-05-15 15:45:16 · answer #1 · answered by steiner1745 7 · 0⤊ 0⤋
Jane is 26
1 year ago she was 25 - perfect square (5^2)
Next year she will be 27 - a perfect cube (3^3)
As a matter of interest 26 is the ONLY integer that lies between a perfect square and a perfect cube. This was proven by Pierre de Fermat (Of Fermat's last theorem fame). I don't think a 6 grader would be required to prove it!
2007-05-15 19:58:04 · answer #2 · answered by welcome news 6 · 0⤊ 0⤋
Jane is 26 years old.
One year ago Jane was 25 years old. 25 = 5^2
One year from now, Jane will be 27. 27 = 3^3
2007-05-15 19:57:03 · answer #3 · answered by A Curious Savage 1 · 0⤊ 0⤋
Jane is 26
2007-05-15 19:58:17 · answer #4 · answered by davidosterberg1 6 · 0⤊ 0⤋
26
26 - 1= 25= 5 squared
26 +1 = 27 = 3 cubed
2007-05-15 19:54:23 · answer #5 · answered by Georgia J 2 · 2⤊ 0⤋