How does one subtract from concatenation in order to prove that $4\times 5 + 67 = 45 + 6\times 7$? I noticed that if we get the numbers $4$, $5$, $6$ and $7$, they have an interesting property!

$$4 \times 5 + 67 = 45 + 6 \times 7\tag*{= 87.}$$

I then conjectured that these were the only four consecutive positive integers with this property.
Hmm...

Let the four integers be $n$, $n+1$, $n+2$ and $n+3$. Then we have $$\begin{align} &n(n+1) + (n+2)\| (n+3) \\ =\;&n\| (n+1) + (n+2)(n+3).\tag{$\|\;\small{ \rm stands\;for}\;concatenation$}\end{align}$$ Or $$\begin{align} (n+2)\|(n+3)-n\|(n+1)&=(n+2)(n+3)-n(n+1) \\ &=n^2 + 5n + 6 - n^2 - n \\ &= 4n+ 6 \\ &= 2(2n+3).\end{align}$$
Now this is when I noticed that if $n=4$ then $2(2n+3)=22$. I may be able to reach $22$ like so:

$$\begin{align}(n+2)\|(n+3)-n\|(n+1)&=(n+2-n)\|(n+3-n-1)\\ &=2\|2 \\ &= 22\end{align}$$

If that is true, then this happens: $$\begin{align}22 &= 2(2n+3)\\ \Leftrightarrow \; 11 &= 2n+3 \\ \Leftrightarrow \; \phantom{1}8 &= 2n \\ \\ \therefore \; n &= 4.\end{align}$$ This is my desired result, thus proving the conjecture. But is the equation in the second sandbox true?


How does one subtract from concatenation?


This, in particular, is new to me; I would mostly appreciate a full, explanatory answer.

Thank you in advance.

We have the following cases:


*

*$n+1,n+3$ both have $k\ge1$ digits; then$$(n+2)||(n+3)-n||(n+1)=10^k(n+2)+(n+3)-10^k(n)-(n+1)\\=2(1+10^k)$$For solutions to your problem in this case,$$2n+3=1+10^k$$You can get infinitely many solutions $\{4,49,499,4999,...\}=\{10^k/2-1:k\in\Bbb N\}$ using different $k$, meaning that the solution is not unique.

*$n+3$ has $k+1$ digits and $n+1$ has $k\ge1$ digits; then $$(n+2)||(n+3)-n||(n+1)=10^{k+1}(n+2)+(n+3)-10^k(n)-(n+1)\\=2+10^k(9n+20)$$The solutions are given by$$2n+3=1+10^{k-1}(45n+100)$$Note that $10^{k-1}(45n+100)\ge45n+100>2n+3$, so there is no solution in this case. 
