Use your imagination!

Problem:

Prove that the sequence $1,11,111,...$ contains no perfect squares other than $1$.

Solution 1:

Notice that every term after the first term is congruent to $11 \equiv 3 \pmod 4$. But no square is congruent to $3 \pmod 4$, so no term after the first term is a perfect square.

Solution 2:

Observe that

$111...11=\frac{999...99}{9}=\frac{10^n-1}{9}$

for some $n \in Z. (n>1)$

Assume that

$\frac{10^n-1}{9}=k^2$.

Then

$10^n-1=9k^2=(3k)^2$

so

$10^n=(3k)^2+1$.

But since $n>1$, by Mihailescu's Theorem, we get a contradiction.

So there are no perfect squares in the sequence except $1$.

Problem 1: Prove that $\sqrt[3]{7}$ is an irrational number.

Conventional solution:

Prove by Rational Root Theorem. Suppose otherwise, then let $x= \sqrt[3]{7}$ for rational $x$, equivalently, we have $x^3 - 7 = 0$. By rational root theorem, we have $\pm 1, \pm 7$ as possible rational solution, however, by trial and error, we can see that none of these values satisfy the equation, which is absurd. Thus, it must be irrational.

Overkill solution:

Prove by Fermat's Last Theorem. Suppose otherwise, then let $\sqrt[3]{7} = \frac pq$ for coprime positive integers $p,q$. Then $7 = \frac{p^3}{q^3} \Rightarrow 7q^3 = p^3$, or $8q^3 = p^3 + q^3 \Rightarrow (2q)^3 = p^3 + q^3$ which contradicts Fermat's Last Theorem.