Persons A and B each have a positive integer number of tokens, and the number of tokens B has is a square number less than 100. B says to A, "If you give me all of your tokens, my total number of tokens will still be a square number." A says, "Yes - if on the other hand, you give me the same number of tokens that I already have, your total number of tokens will also be a square number." How many tokens does A have?
2026-04-01 08:04:16.1775030656
On
How many tokens would person A have under these conditions?
134 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
2
There are 2 best solutions below
1
On
Well, B has $25$, and A has $24$, and there are no other solutions.
I know this is not a proof though.
Here is some sort of proof:
1) Write the numbers: $1,4,9,16,25,36,49,64,81$
2) If A has $n^2$, and B has $m$ then
$n^2 - m = x^2 $
$n^2 + m = y^2 $
Hence, $2n^2 = x^2 + y^2$
This means that $x^2, n^2, y^2$ form an arithmetic progression i.e.
$n^2-x^2 = y^2-n^2$
Now, upon a closer look at 1)
(we can write down all $9.8/2$ pairwise differences and check them),
we can see this happens to be true only when $x=1, n=5, y=7$.
This gives us the unique solution mentioned above.
For a more detailed study see:
Let's start by listing all the square numbers between $0$ and $100$ $$1,4,9,16,25,36,49,64,81$$ Note we do not include $100$. We know that $$B \in [1,4,9,16,25,36,49,64,81]\\B+A \in [1,4,9,16,25,36,49,64,81,100,121,144,...]\\ B-A \in [1,4,9,16,25,36,49,64,81,100,121,144,...]$$ So what we need to find in a square equidistant from 2 other squares. Note that the last 2 sets aren't limited to be less than 100. We can derive some more equalities $$B= b^2, A=a, B+A=c^2, B-A=d^2\\b^2+a=c^2\\b^2-a=d^2\\b^2+b^2=c^2+d^2\\2b^2=c^2+d^2$$ This tells us that the sum of our two new squares can't be more than twice $b^2$. At this point we can almost brute force it as we only have $9$ choices for $b$. going through the list, $b$ cannot be $1,2,3,4$. When we get to $b=5$ however, we have $$2(5)^2= c^2 + d^2\\50 = c^2+d^2\\50=1^2+7^2\\50=1+49\\50=50$$So the solution to our problem is $$A=24$$