Let $Q(n)$ (from German Quersumme) denote the sum of the (base-ten) digits in the integer $n$.
It is a well-known fact that $Q(n)$ is congruent modulo 9 to $n$.
Also, if you look down the diagonal of the base-9 multiplication table, you'll see that all the perfect squares end in 0, 1, 4, 7. Put another way, if $n$ is a perfect square, then:
$$Q(n) \equiv n \in \{0, 1, 4, 7\} \pmod{9}$$
For both $Q(n)$ and $Q(n+1)$ to be perfect squares, the only possible combination is $Q(n) \equiv 0 \pmod{9}$, so that $Q(n+1) \equiv 1 \pmod{9}$.
Now, let's choose two perfect squares, one of which is a multiple of 9, and one that's one more than a multiple of 9. I'll pick $Q(n) = 81$ and $Q(n + 1) = 64$.
Note that in order for $Q$ to decrease when $n$ is incremented, we need a rollover from 9 to 0 in the unit's place. So, we have:
- $n$ = (99 digits adding to 72)$9$
- $n + 1$ = (99 digits adding to 64)$0$
Since $72 > 63$, we'll need another place-value column affected by rollover.
- $n$ = (98 digits adding to 63)$99$
- $n + 1$ = (98 digits adding to 64)$00$
OK, so now we just need two consecutive digits in the hundred's place. I'll pick $8$ and $9$.
- $n$ = (97 digits adding to 55)$899$
- $n + 1$ = (97 digits adding to 55)$900$
And the “97 digits adding to 55” must be the same string of digits for the two numbers. I'll choose to construct it as 1
+ 90 0
's + 6 9
's. giving the pair of numbers:
- 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000999999899
- 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000999999900
So there you go: Two consecutive 100-digit integers whose digit sums (81 and 64) are both perfect squares.