"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?

2. 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

3. Originally Posted by Mukilab
"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
No, either x and y are both even or x and y are both odd.

Now I'm stuck

any help?
I think the best way to do this is just by "brute strength".

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.

4. 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.

5. 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