Squares Ending in 5 Occurrences of 2-Digit Pattern

Theorem
Let $n$ be a square number whose decimal representation ends in the pattern $\mathtt {xyxyxyxyxy}$.

Then $\mathtt {xy}$ is one of:
 * $21, 29, 61, 69, 84$

The smallest examples of such numbers are:

Case $1$: $\mathtt {xy}$ is odd
Consider the last $3$ digits of $n$: $\mathtt {yxy}$.

By Odd Square Modulo 8:
 * $n \equiv \mathtt {yxy} \equiv 1 \pmod 8$

By Square Modulo 5:
 * $n \equiv \mathtt {yxy} \equiv 0, 1, 4 \pmod 5$

We have:

Since $y$ is odd, we have $\mathtt {yxy} \equiv 1 \mathtt {xy} \pmod {200}$.

Hence $\mathtt {xy}$ can only be $05, 21, 29, 45, 61, 69, 85$.

Suppose $n$ ends in $5$ in decimal representation.

By Divisibility by 5, $5 \divides n$.

Since $5$ is a prime, $5 \divides \sqrt n$.

By Odd Square Modulo 8 and Chinese Remainder Theorem:
 * $n \equiv 25 \pmod {200}$

for which none of $\mathtt {xy} = 05, 45, 85$ satisfy.

This shows that $\mathtt {xy}$ can only be $21, 29, 61, 69$.

Case $2$: $\mathtt {xy}$ is even
Consider the last $2$ digits of $n$: $\mathtt {xy}$.

We omit the trivial case $\mathtt {xy} = 00$.

This case occurs whenever $10 \divides \sqrt n$.

Since $\sqrt n$ is even:
 * $n \equiv \mathtt {xy} \equiv 0 \pmod 4$

Since $5 \nmid \sqrt n$, by Square Modulo 5:
 * $n \equiv \mathtt {xy} \equiv 1, 4 \pmod 5$

We have:

In the list above:
 * $4, 24, 36, 44, 56, 76, 84$ are divisible by $2^2$ but not $2^4$
 * $16, 96$ are divisible by $2^4$ but not $2^6$
 * $64$ is divisible by $2^6$

Write $n = 10^{10} k + 101010101 \times \mathtt {xy}$.

We require $\dfrac n {2^{2 d}}$ to be a square number as well, where $d$ is the largest integer where $2^{2d} \divides \mathtt {xy}$ (shown above).

We have:

and:
 * $\mathtt {xy} = 04, 16, 24, 36, 44, 56, 64, 76, 84, 96$

corresponds to:

Thus the only valid $\mathtt {xy}$ is $84$.

We have considered all possibilities, and the only possible $\mathtt {xy}$ are:
 * $21, 29, 61, 69, 84$

and by the examples shown, all these numbers do indeed produce squares.