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Problem 6

Let x, y, and z be positive integers such that x is a two digit number and y is obtained by reversing
the two digits of x. The integers satisfy the equation x² - y² = z².
Find the ordered pair (x,y,z).

Solution

Let m and n represent the digits of x and y, such that x=10m+n and y=10n+m.
By the difference of squares, x²-y²=(x+y)(x-y)=(10m+n+10n+m)(10m+n-10n-m)=(11(m+n))(9(m-n))
For the product to be a square, the factor of 11 must be repeated in either (m+n) or (m-n), and given that
m and n are digits, they must be positive, and thus only (m+n)=11 can be satisfied. Since 9 is a square, it
can be ignored, and thus (m-n) must also be a square.
Since (m-n) must be a square, and m cannot be 10 or greater, (m-n) must equal 4 or 1.
If m-n=4, then (m+n)+(m-n)=11+4, and m=15/2 which is not a digit.
If m-n=1, then (m+n)+(m-n)=11+1, so m=6 and n=5. Therefore x=65 and y=56.
Solving the original equation with x and y known, we find that z=33.
Therefore the ordered pair is (65,56,33).