Show that no cube of an integer can be expressed as $7n + 5$ for some positive integer $n$
This is from Riley's "Mathematical methods for Physics and Engineering", and is question 1.28 b, from the section "proof by induction and contradiction"
Attempt: have formed the equation $$ x^3 = 7n + 5$$ and rearranged to get $$ \frac{x^3 - 5}{7} = n $$
so I have the condition that $x^3 - 5$ must be divisible by $7$, for $n$ to be an integer.
For the proof by induction:
Hint: Notice that $(x-7)^3 = x^3 - 3 × 7 x^2 + 3 × 7^2 x - 7^3$.
This means if $x^3 = 7n+5$ then $ (x-7)^3$ is also $7n+5$ for some n. Because $-3 × 7 x^2 + 3 ×7^2 x - 7^3 = 7 (-3 x^2 + 3 × 7 x - 7^2)$ $7 × (some ~integer)$
Although this is probably not what you're looking for, a simpler proof would be:
You could also treat it as $x^3 \equiv 5$ (mod 7). Then using Fermat's Little Theorem: $x^6-1 \equiv 0$ so $(x^3)^2-1 \equiv (x^3 -1) (x^3 +1) \equiv 0 $ (mod 7). So $x^3$ modulo 7 is in general, -1,0 or 1.
