an (integer) solution is (78,64), but i could only get it with a spreadsheet. There are an unlimited number of solutions in real numbers that you could get from the quadratic formula
"In 1988 my Grannie was quite a bit older than Grandpa. In fact the difference between the squares of their ages (amazingly!) was exactly 1988.... How old were they?"
My workings:
x^2-y^2=1988
(x+y)(x-y)=1988
(x+y) or (x-y) must be even
x or y must be even
Now I'm stuck
any help?
No, either x and y are both even or x and y are both odd.
I think the best way to do this is just by "brute strength".Now I'm stuck
any help?
Lets just look at some possible numbers. Lets start With Grannie being, say 80. Then 80^2- x^2= 6400- x^2= 1988 so x^2= 4412. The positive root is 66.4.
80^2- 66.4^2= 1991- a little too large. Grannie and Grandpa must be closer together in age.
If Grannie were 75, we would have 75^2- x^2= 5625- x^2= 1988 so x^2= 3637 and Grandpa must be 60.3. Now, 75^2- 60.3^2= 1988.81.
You can try some ages around 75 to see if you can come closer.
I found an easier way to solve it.
If you look for the difference of two squares and make a pattern, you will see that it has to be in the form of
4 (n+1) 3(2n+3)
8(n+2) 5(2n+5)
12(n+3) 7(2n+7)
16(n+4) 9(2n+9)
20(n+5)
24(n+6)
28(n+7)
If you factor 1988 you get 4, 7, 71 , 28 times n+7 = 71 then n=64
to get 28(n+7) means you are looking at the difference of n squared and (n+14) squared.
Thanks Veronica for that different method but I still think brute force would be easier. Once I had eliminated the odd integers for x and y I went from 70 (since it's a grandmother) and did it in less than 20 seconds quickly substituting the integers on my calculator.
Thanks for the help guys