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
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.
This leads to the equation
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 =
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
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