I think what you are asking is, can you have a series of 3 consecutive squares which sums to a square; or 4, or 5, etc.
It seems there are an infinite number of series of two consecutive squares which sum to a square.
(a,a+1,b) --> a^2 + (a+1)^2 = b^2
(0,1,1)
(3,4,5)
(20,21,29)
(119,120,169)
(696,697,985)
See Sloane's A008844
(http://oeis.org/A008844)
So I think if the above were the only ones, then the answer to your question would be "two".
But your hint says there is a two digit number, so I looked a little further.
It works for 11 (but I can't prove that is the only one)
Sum of squares(18,19,...,28) = 77^2
Sum of squares(38,39,...,48) = 143^2
Sum of squares(456,...,466) = 1529^2
Sum of squares(854,...,864) = 2849^2
Sloane A218395 (http://oeis.org/A218395)
A curiosity is that the Sloane series (11,77,143,...) begins with 11 not 77.
It turns out that:
Sum of squares(-6,...,+4) = 11^2
Your hint was helpful
