# What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$?

For a positive integer $$n$$ and nonzero digits $$a$$, $$b$$, and $$c$$, let $$A_n$$ be the $$n$$-digit integer each of whose digits is equal to $$a$$; let $$B_n$$ be the $$n$$-digit integer each of whose digits is equal to $$b$$, and let $$C_n$$ be the $$2n$$-digit (not $$n$$-digit) integer each of whose digits is equal to $$c$$. What is the greatest possible value of $$a + b + c$$ for which there are at least two values of $$n$$ such that $$C_n - B_n = A_n^2$$?

What I have tried:

I start by trying $$n = 1$$ and $$n = 2$$. These give the system of equations $$11c - b = a^2$$ and $$1111c - 11b = (11a)^2$$. These imply that $$a^2 = 9c$$, so the possible $$(a, c)$$ pairs are $$(9, 9)$$, $$(6, 4)$$, and $$(3, 1)$$. The first puts $$b$$ out of range but the second makes $$b = 8$$. We now know the answer is at least $$6 + 8 + 4 = 18$$.

How should I continue?

Help is appreciated! Also, if you are nice, can you PLEASE help me on this question($N$'s base-5 and base-6 representations, treated as base-10, yield sum $S$. For which $N$ are $S$'s rightmost two digits the same as $2N$'s?)?

Thanks!

Max0815

• How do you get from your first two equations to $a^2=9c?$ You can divide the second by $11$ to get $101c-b=11a^2$ Commented Feb 5, 2019 at 23:53
• I'm not sure what you mean by your first equation. And how does dividing by 11 help though... :( I'm confused Commented Feb 5, 2019 at 23:55
• You said correctly that for $n=1$ you have $11c-b=a^2$ and for $n=2$ you have $1111c-11b=(11a)^2$. I don't see how either of those leads to $a^2=9c$. I divided the $n=2$ equation by $11$ to get my result, which I think is a simpler equation to work with. Commented Feb 5, 2019 at 23:59
• Right... ahh I think I had an algebra error. Let me try again with ur equation... Commented Feb 6, 2019 at 0:02
• @RossMillikan Now subtract $11c - b = a^2$ from $101c - b = 11a^2$. Doesn't it give $90c = 10a^2$? Commented Feb 6, 2019 at 0:03

We note that $$A_n = a\cdot\dfrac{10^n - 1}{9}$$, $$B_n = b\cdot\dfrac{10^n - 1}{9}$$, and $$C_n = c\cdot\dfrac{10^{2n} - 1}{9}$$. Then, we have $$c\cdot\dfrac{10^{2n} - 1}{9} - b\cdot\dfrac{10^n - 1}{9} = (a\cdot\dfrac{10^n - 1}{9})^2$$, and when we multiply everything out and simplify, we get $$9c\cdot10^n + 9c - 9b = a^2 \cdot 10^n - a^2 \implies (9c-a^2) \cdot 10^n + 9c-9b + a^2 = 0$$ Now, note that since if two distinct values of $$n$$, say $$x$$ and $$y$$ work, then we get the equations$$(9c-a^2) \cdot 10^x + 9c-9b + a^2 = 0$$ and $$(9c-a^2) \cdot 10^y + 9c-9b + a^2 = 0$$, and subtracting gets $$(9c-a^2) \cdot (10^y - 10^x) = 0$$, and since $$10^y - 10^x \neq 0$$, we must have $$9c = a^2$$. Therefore, we get three solutions for $$(a,c)$$, which only $$(6,4)$$ gives an integer value of $$b$$, which means the solution $$(a,b,c)$$ is $$(6,8,4) \implies 18$$.
You were on the right track for finding $$a^2 = 9c$$, however, the problem does become easier after converting to $$10^n$$ instead of dealing with increasingly larger numbers; another way was to prove that $$(a,c) = (3,1)$$ is too small even if $$b = 9$$, and that $$(a,c) = (9,9)$$ has no solutions by noting that $$C_n - (A_n)^2 = B_n$$ and that $$99 - 9^2 = 9(11-9) = 18$$, $$9999 - (99)^2 = 99(101-99) = 99 \cdot 2$$, and that $$9_n - (9_n)^2 = 2 \cdot 9_n$$. (This can be proved by realizing that $$C_n = A_n \cdot (A_n + 2)$$.) Since $$2 \cdot 9_n$$ can never be equal to $$B_n$$, as $$2 \cdot 9_n$$ has a units digit of $$8$$ but it's first digit is $$1$$, we are done.