SOLUTION: A two digit number has different digits. If the difference between the square of the number and the square of the number whose digits are interchanged is a positive perfect square,

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Question 252191: A two digit number has different digits. If the difference between the square of the number and the square of the number whose digits are interchanged is a positive perfect square, what is the two digit number?
Found 2 solutions by drk, Edwin McCravy:
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
Let 10T + U be the first digit, where T can't = U.
Let 10U + T be the second, reversed digits, number, where T can't = U.
From the question, we then get,
%2810T%2BU%29%5E2+-+%2810U%2BT%29%5E2+=+X%5E2
100T%5E2+%2B+20UT+%2B+U%5E2+-100U%5E2+-+20UT+-+T%5E2+=+x%5E2
we can see that the UT parts cancel. After combining like terms, this leaves us with
99T%5E2+-+99U%5E2+=+X%5E2
99%28T%5E2-U%5E2%29 = X^2}}}
3%2Asqrt%28%2811%29%2A%28T%2BU%29%28T-U%29%29+=+X%5E2
SInce 6+5 = 11 and 6-5 = 1, T = 6 and U = 5, we get
65^2 - 56^2 = 4225 - 3136 = 1089 = 33^2.

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

This time drk's solution was shorter than mine.  Essentially he
took square roots of both sides of

99%28t-u%29%28t%2Bu%29=k%5E2

getting

3sqrt%2811%2A%28t-u%29%28t%2Bu%29%29=k

Ssince the right side is a positive integer, the left side must 
be also.  That means there must be another factor of 11 under 
the square root radical, and the only possibility of the sum of 
two digits being a multiple of 11 is for (t+u) to be 11, and that 
is only true when t=6 and t=5. 
 
Here was my longer solution:
-----------------------------

A two digit number has different digits. If the difference between the square of the number and the square of the number whose digits are interchanged is a positive perfect square, what is the two digit number? 


There must exist integer k so that:

%2810t%2Bu%29%5E2-%2810u%2Bt%29%5E2=k%5E2

Factor the left side as the difference of two perfect squares:

%22%5B%22%2810t%2Bu%29-%2810u%2Bt%29%22%5D%22%22%5B%22%2810t%2Bu%29%2B%2810u%2Bt%29%22%5D%22%22=%22k%5E2

Simplifying:

%22%5B%2210t%2Bu-10u-t%22%5D%22%22%5B%2210t%2Bu%2B10u%2Bt%22%5D%22%22=%22k%5E2

%289t-9u%29%2811t%2B11u%29=k%5E2

9%28t-u%29%2A11%28t%2Bu%29=k%5E2

99%28t-u%29%28t%2Bu%29=k%5E2

%28t-u%29%28t%2Bu%29=%28k%5E2%29%2F99

Write 99 as a product of primes:

%28t-u%29%28t%2Bu%29=%28k%5E2%29%2F%283%5E2%2A11%29

Since the left side is an integer, the right side must be also.
So k%5E2 must contain factors 3%5E2%2A11%5E2 as well as 
possibly another perfect square m%5E2.

So k%5E2 must be a perfect square of the form 3%5E2%2A11%5E2%2Am%5E2,
where m is a positive integer. 

%28t-u%29%28t%2Bu%29=%283%5E2%2A11%5E2%2Am%5E2%29%2F%283%5E2%2A11%29

That gives:

%28t-u%29%28t%2Bu%29=11m%5E2

The largest possible value of (t-u)(t+u) is 
when t=9 and u=0, or (9-0)(9+0)=81 and the smallest
is when t=1 and u=0, or (1-0)(1+0)=1.  Therefore the
right side must satisfy this:

1%3C=11m%5E2%3C=81
1%2F11%3C=m%5E2%3C=81%2F11=7%264%2F11
1%3C=m%5E2%3C=4 since 4 is the largest perfect 
square that does not exceed 7%264%2F11, and since
1 is the smallest largest perfect square that exceeds
1%2F11.
Therefore
1%3C=m%3C=2

So m is either 1 or 2.

When m = 2, we have

%28t-u%29%28t%2Bu%29=44

The only ways 44 can be written as the product of two
integers are 1*44, 2*22, and 4*11. But the larger factor,
t+u cannot be more than 9+8 or 17, thus that would leave
only 4*11.  4 is even and 11 is odd. However t-u and t+u 
are either both even or both odd.  So we have ruled out 
m = 2.

That leaves only the possibility 

m = 1 and

%28t-u%29%28t%2Bu%29=11

Since 11 is prime, the only two positive integer factors it has
are 1 and 11.  Therefore the smaller factor %28t-u%29 must be 1
and the larger factor 11.  So we have the system of equations:

system%28t-u=1%2Ct%2Bu=11%29

Solving that gives us t=6 and u=5, which both can be digits.  
So the only solution is the two digit number 65.

Edwin