# Finding every solution such that ${kk\cdots k}^2+ll\cdots l=mm\cdots m$ such as ${333}^2+222=111111$

Let $[k,n]$ be a $n$-digit integer such that $kk\cdots kk$ in the decimal system where $k=1,2,\cdots,9$. For example, $[8,2]=88, [8,5]=88888$.

Then, here is my question.

Question 1 : Can we find every $(a,b,c,d,e,f)\ (a,b,c,d,e,f\in\mathbb N, 1\le a,c,e\le 9)$ such that $${[a,b]}^2+[c,d]=[e,f]\ ?$$

Motivation : I've known the following equations :

$$3^2+2=11, {33}^2+22=1111, {333}^2+222=111111,\cdots$$ $$6^2+8=44, {66}^2+88=4444, {666}^2+888=444444,\cdots$$ We can prove that $(a,b,c,d,e,f)=(3,n,2,n,1,2n),(6,n,8,n,4,2n)$ are sufficient for any $n\in\mathbb N$. These led me to the above question, but I'm facing difficulty for finding every possible solution. Can anyone help?

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Apart from the trivial one-digit solutions $1^2 + l = 1+l,\: 1\leqslant l \leqslant 8$, $2^2 + l, \: l \in \{1,2,3,4,5,7\}$, $4^2 + 6 = 22$, $5^2 + 8 = 33$, $7^2 + 6 = 55$, $8^2 + 2 = 66$, $9^2+7 = 88$, and the regular solutions

\begin{align} \left(3\frac{10^k-1}{9}\right)^2 + 2\frac{10^k-1}{9} &= \frac{10^{2k}-1}{9}\\ \left(6\frac{10^k-1}{9}\right)^2 + 8\frac{10^k-1}{9} &= 4\frac{10^{2k}-1}{9}, \end{align}

there is only the mildly interesting

$$88^2 + 33 = 7777.$$

For the squares with a base greater than $10$, we have

\begin{align} 11^2 &\equiv 21 \pmod{100}\\ 22^2 &\equiv 84 \pmod{100}\\ 33^2 &\equiv 89 \pmod{100}\\ 44^2 &\equiv 36 \pmod{100}\\ 55^2 &\equiv 25 \pmod{100}\\ 66^2 &\equiv 56 \pmod{100}\\ 77^2 &\equiv 29 \pmod{100}\\ 88^2 &\equiv 44 \pmod{100}\\ 99^2 &\equiv 01 \pmod{100}. \end{align}

It is easily checked that adding a single digit to the square never produces a number of the form $d\frac{10^k-1}{9}$, and for numbers with more than one digit, a necessary condition is that $aa^2 + cc \equiv e\cdot 11 \pmod{100}$, which for the remainder of the square modulo $100$ means that it must be a multiple of $11$, or $100$ minus a multiple of $11$, which is a multiple of $11$ plus $1$. If $aa^2 \equiv r\cdot 11 \pmod{100}$, then we have $aa^2 + bb \equiv (r+b)\cdot 11 \pmod{100}$ for $1 \leqslant b \leqslant 9-r$, so these are candidates. If $aa^2 \equiv r\cdot 11+1 \equiv 100-(9-r)\cdot 11 \pmod{100}$, then we have $aa^2 + bb \equiv (b+r-9)\cdot 11 \pmod{100}$ for $10-r \leqslant b \leqslant 9$, so these are candidates too. For all other remainders, adding a multiple of $11$ between $11$ and $99$ (inclusive) never produces two equal last digits, so these aren't candidates.

The only numbers satisfying these requirements are $88$, but $888^2 \equiv 544 \pmod{1000}$ (and the first digit is $7$, so adding $11$ doesn't produce a number of the desired form), so $88$ does not generate a sequence, $33$ and $66$, which two generate a sequence.

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Thank you very much for your answer with nice ideas. I would like you teach me more about the part that "a multiple of $11$ plus $1$". I'm confused about if ${33}^2\equiv 89$ satisfies ${aa}^2+cc\equiv e\cdot 11$. – mathlove Nov 17 '13 at 5:11
The point is that "multiple of $11$ plus $1$" is "$100 - m\cdot 11$", so adding $(m+1)\cdot 11,\, \dotsc,\,99$ produces two equal last digits. I've elaborated that a bit in the answer, hope that clarifies it. – Daniel Fischer Nov 17 '13 at 10:58
Yes, that does clarify it. Thank you very much. – mathlove Nov 17 '13 at 11:01