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