Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2
What is wrong with these dodgy proofs?
To find all 3-digit numbers $abc$ (in base $10$) such that
$$ a + b^2 + c^3 = 100a + 10b + c $$ Rearranging gives $$c^3 - c
- 99a = b (10 - b)$$ $$c(c+1)(c-1) - 99a = b (10 - b)$$
For any three consecutive integers one of them is divisible by
$3$. Since $3$ divides $99$ it follows that $3$ divides $b(10-b)$.
Since $3$ is a prime this limits the possible choices of $b$:
Hence the possible values of $b(10-b)$ are $0$, $9$, $21$ and
We now have to find a multiple of $99$ which when subtracted
from a product of $3$ consecutive natural numbers gives $0$, $9$,
$21$ or $24$.
Since $a$ is at least $1$ $c(c+1)(c-1)$ is at least $99$ so $c$ is
at least $5$.
(Since $0$, $9$, $21$, $24 < 99$ only the multiple of $99$
which is closest to $(c+1)c(c-1)$ needs to be checked.)
From the table we can see that the following are the
possibilities for $a$, $b$ and $c$:
$a=1$ $b=3$ or $b=7$ $c=5$
$a=5$ $b=1$ or $b=9$ $c=8$
giving the solutions $135$, $175$, $518$ and $598$.