Very squary number

For every non-zero digit, there is at most one possible next digit. The complete set of possibilities is $8 \to 1 \to 6 \to 4 \to 9; \ 2 \to 5; \ 3 \to 6.$ 7 does not occur in any two-digit perfect square. Since we do not start a number with zero and no perfect square between 0 and 100 ends in zero, we can also rule out the use of zeroes.

Thus the longest number satisfying the condition is $\boxed{81649}$ .

The two-digit perfect squares are $16, 25, 36, 49, 64, 81$ . Consider

$16 - 64 - 49$ giving us $1649$ .

$25$ .No two digit square starts with $5$ so this gives $25$

$36 - 64 - 49$ , we get $3649$ .

$49$ .

$64 - 49,$ gives $649$ .

$81 - 16 - 64 - 49$ gives $81649$ .

So the greatest of all is $\boxed{81649}$

For the number to be greater, the first digit on the left-hand side should be the greatest. So, we get 9. But there is no two digit perfect square with ten's digit as 9. So, we go down to 8. We have one such perfect square - 81. So, from the left-hand side, our first two digits are 8 and 1. Now, from 1, the only number is 16. So, our number becomes 816... From 6, we get 64. So, our number becomes 8164... From 4, we get 49. So, our number becomes 81649... But, from 9, there is no perfect square and thus, we stop. Our final answer is 81649.

nice solution