Do you have any problems that can be solved very easily with an alternative solution using very high-level theorems? I have added one below.

Note by Shenal Kotuwewatta
3 years ago

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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.

- 3 years ago

Kind of similar, so adding this as a comment.

* Problem 2: * Prove that $$\sqrt[n]{2}$$ is an irrational number for $$n > 2$$.

* Overkill solution: *

Proof by Fermat's last theorem. Suppose otherwise, then let $$\sqrt[n]{2} = \frac{p}{q} \implies 2 = \frac{p^n}{q^n}$$, or $$2q^n = p^n \implies q^n + q^n = p^n$$, which contradicts Fermat's last theorem.

Hey, did you stumble upon this Wiki by accident?

- 3 years ago

Nope. What a coincidence. xD

I don't really use the wikis. I did see this somewhere on Brilliant though. Much before the wikis. Maybe it was one of Sreejato's questions. I'm not sure TBH.

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 $1$.

- 3 years ago