Following Sudipta Das (see the Ed Pegg's page at
mathpuzzle.com/)
we define the n-complement of a number j as the number k
such that corresponding digits of j and k always add to n.
We want to investigate some properties of the n-complements in two special cases
for the terms j, k:

the square-case, say
j=x^{2} and
k=y^{2} for some x, y;

the triangular-case, say
j=x*(x+1)/2, k=y*(y+1)/2 for some x, y.

Let us sketch an idea that apply to the general problem (not only to our two cases).
Forbidding initial zeros on the representation of our numbers, we assume that
j, k have the same numbers of digits, say d; then we must have:

Main Formula: j+k=n*S where
S=111...1 is composed by d times the digit 1.

Of course carries could appear in the sum j+k, thus solutions of the new
problem could fail to solve the original one; but for the new formulation many
interesting properties will be quite immediate.

Square-Complements

Some trivial cases

Of course any choice of x, y between 0 and 3 will give rise to a solution
corresponding to a suitable n; thus we get trivial solutions for 10 values of
n = x^{2}+ y^{2}
say (choosing x £ y and sorting the results
with respect to n):

x

0

0

1

2

1

2

3

1

2

3

y

0

1

1

2

2

2

3

3

3

3

n

0

1

2

4

5

8

9

10

13

18

Of course we could do the same work concerning the two-digit-cases (say: x, y
both between 4 and 9), but it would be a quite tedious work, and a better idea could be
to ask the computer to do that. In fact, as we will show in a moment, we would waste our
time: no two-digits solution can exist.

A little bit of theory can save a lot of time...

Let us start from the "main formula" (in our case, say with
j=x^{2}, k=y^{2})
and remark that, once d has been fixed, we can easely find some factors of the
part S (the parts "thus n*S has the factor..." will be
explained in a moment):

if d is even, the number S has the factor 11
(thus n*S has the factor 11*11 for n other than 11);

if d is a multiple of 3, the number S has the factor
3 (thus n*S has the factor 3*3 if 3 does not divide
n);

if d is a multiple of 5, the number S has the factor
271 (thus n*S has the factor 271*271);

if d is a multiple of 6, the number S has the factor
7
(thus n*S has the factor7*7 if 7 does not divide
n);

if d is a multiple of 7, the number S has the factor
239 (thus n*S has the factor 239*239);

if d is a multiple of 13, the number S has the factor
79 (thus n*S has the factor 79*79);

....

Let us explain the special role played by the numbers 11, 3, 271, 7, 239, 79.
In general there is no relationship between the factors of x, y and the factors
of x^{2}+ y^{2}
(consider e.g. the example
7^{2}+5^{2} =
74 = 2*37); however in some cases a very strong connection exists: special
factors of the sum must divide both x, y; so that also the square
of the factor divides the sum. This happens in particular for each one of the values
11, 3, 271, 7, 239, 79; of course, with respect to the representation given
in the main formula, some terms could divide n instead of the string of ones.