Problems & Puzzles: Puzzles

 Puzzle 720 x^2&y^2 or y^2&x^2 squares żAre there pairs of consecutive integers (x>3, y>3) such that x^2&y^2 or y^2&x^2 (& means concatenation) are squares? The only known examples are: (2,3) -> 49 (3,4)-> 169 Q Are there more examples or they are impossible?

Contributions came from Emmanuel Vantieghem

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Emmanuel wrote:

About puzzle 720, I think there are no more solutions.  I do not have a proof, but I have some arguments that make my guess plausible.

Take some natural number  k > 0. When we look for numbers  m  with the property
m^2 + 10^k *(m+or-1)^2 = y^2   (**)
then we can find infinitely many solutions (which will be sollutions of puzzle 720 iff  k  is the number of digits of  m^2).
Indeed, the equation can be written in the form
m^2 (1+10^k)+/-2m*10^k+10^k-y^2 = 0.
Whence, the 'discriminant'(divided by 4)  10^(2k) - (1+10^k)(10^k - y^2)  should be a square, say  x^2.
x^2 - y^2 (1+k0^k) = -10^k    (***)
Which has the Obvious (positive) solution  x = 1, y = 1.
To find other solutions, we just have to solve the Pell-equation
u^2 - v^2 (1+10^k) = 1
which has infinitely many solutions.  All of them yield infinitely many solutions of (***)
(just take  x = u+v(1+10^k)  and  y = u+v ).
For instance, if  k = 2n  is even, then a solution of the corresponding Pell equation is
u = 2*10^2n + 1 , v = 2*10^n
This will give as smallest solutions of  (**) :
k = 2  -> 21^2 + 100*22^2 = 221^2
k = 4  -> 201^2 + 10000*202^2 = 20201^2
k = 6  -> 2001^2 + 1000000*2002^2 = 2002001^2
...
Eilas ! These solutions are not solutions of puzzle 720 for in all cases, m^2  has too many digits !
When  k  is odd >2  the solution of the corresponding pell-equation turns out to be much bigger.
Of course, the weak point in my argumentation is that I cannot prove that there are no other solutions ..
Also, I forgot to mention that one also can take  y = u - v  to get a value for  m (but which is also too big to be a solution for puzzle 720).

Also, I forgot to mention that one also can take  y = u - v  to get a value for  m (but which is also too big to be a solution for puzzle 720).

Maybe it is worthwhile to add that the remark at the end of my previous mail also give the following examples :

19^2 + 100*18^2 = 181^2
199^2 + 10000*198^2 = 19801^2
1999^2+1000000*1998^2 = 1998001^2

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