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Fair and Square (Posted on 2003-11-03) Difficulty: 3 of 5
Jack and Jill each have marble collections. The number in Jack's collection in a square number.

Jack says to Jill, "If you give me all your marbles I'll still have a square number." Jill replies, "Or, if you gave me the number in my collection you would still be left left with an even square."

What is the fewest number of marbles Jack could have?

See The Solution Submitted by Ravi Raja    
Rating: 2.3333 (3 votes)

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Solution Lost their marbles (full solution) | Comment 8 of 25 |
I'll use the notation already suggested; Jack has x marbles and Jill has y. Then, x is a perfect square, x+y is a perfect square, and x-y is a perfect square.

The simplest solution is that Jill has zero marbles (y=0), then Jack could have any square number. Even more simply, maybe Jack doesn't have any at all either. Or he has just one, or four.. whatever.

Assume, then, Jill's collection actually contains some marbles.

The number in Jack's collection, is the square of some number, let's say n (n² = x).
The difference of the numbers in their respective collections is the square of some other number, say m (m² = x-y).
The sum, similarly, is the square of another number p (p² = x+y).

We know y is a positive integer. Thus, m<n, so we can say m=n-a for some a>0. Also, p>n, so p=n+b for some p>0.

Then, some algebra:
x-y = m² = (n-a)²

x-y = n² - 2na + a²
x-y = x - 2na + a²
y = 2na - a²

x+y = p² = (n+b)²
x+y = n² + 2nb + b²
x+y = x + 2nb + b²
y = 2nb + b²


Then:
2nb + b² = 2na - a²

2na - 2nb = b² + a²
n = (b² + a²)/2(a-b)


Therefore,
0 < b < a (as could be intuitively determined from fact that the squares of n-a and n+b are in arithmetic progression).

Any values of a and b will work in this equation, as long as n works out to an integer. Since the denominator contains a two, the numerator of the fraction must be even to yield an integral value of a. That is, the sum of a² and b² must be even, so a² and b² are both even or both odd. In either of these cases, their difference will be even as well, so the denominator will be the product of two even numbers, and the numerator must be a multiple of four. Therefore, a and b must both be even for n to work out to an integer. The simplest case for this (0 < b < a, given a and b both even)occurs when b=2 and a = 4.
Thus:
n = (2² + 4²)/2(4-2)
n = (4 + 16)/2(4-2)
n = 20/4
n = 5


So, n = 5, and x = n², the number of marbles in Jack's collection, is 25.

For completeness:
y = 2nb + b² = 2na - a²

y = 2(5)2 + 2² = 2(5)4 - 4²
y = 20 + 4 = 40 - 16
y = 24


And to check:
25 - 24 = 1 is the square of 5-4=1
and
25 + 24 = 49 is the square of 5+2=7.
Edited on November 3, 2003, 12:42 pm
  Posted by DJ on 2003-11-03 12:37:34
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