"Too simple to be true"

Solution 1:

My favorite is the following:

Let $\pi$ be rational, and write $\pi = a/b$ in lowest term. Let $p \neq 2$ be a prime not dividing $a$. Then in $\Bbb{Q}_{p}$, we have

$$ 0 = \sin(pb\pi) = \sin(pa) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!}(pa)^{2n+1} \equiv pa \ (\mathrm{mod} \ p^2), $$

which is absurd since $a \not\equiv 0$ mod $p$. Therefore $\pi$ is irrational.

The essential gap in this too-good-to-be a proof is that a $p$-adic power series may not converge to the same value as in the real field case, even the series consists of only rational terms. Thus the value of $\sin x$ need not coincide in $\Bbb{R}$ and $\Bbb{Q}_{p}$.

This false proof appears in Neal Koblitz's p-adic Numbers, p-adic Analysis, and Zeta-Fnctions.

Solution 2:

Let $\displaystyle \int$ denote $\displaystyle \int_0^x (\cdot ) dx$. Consider solving the equation $$\int f = f-1.$$ Rearranging, we get that $$f - \int f = 1 \implies \left(1 -\int \right)f = 1$$ Hence, $$f = \dfrac1{1 - \displaystyle\int} = \left(1 + \int + \int \int + \int \int \int + \cdots \right)1\\ = 1 + \int_0^x 1 dx + \int_0^x \int_0^x 1 dx + \int_0^x \int_0^x \int_0^x 1 dx + \cdots = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots = e^x$$which indeed satisfies the equation.

Adapted from this post. The post has lot of other interesting answers as well.

Solution 3:

What I have in my mind is Wilson's theorem, which says that if $p$ is a prime number, then $$(p-1)!\equiv -1 \pmod p.$$

The fake proof I have learned is the following: Since $p=p-1+1$, by taking factorial on both sides, we have $$p!=(p-1+1)!=(p-1)!+1!=(p-1)!+1.$$ Now taking mod $p$, we obtain $$0\equiv(p-1)!+1 \pmod p.$$

Of course the "proof" is wrong. The gap occurs because factorial is not distributing in the sense that $(a+b)!\neq a!+b!$ in general. In fact, same "proof" would work without assuming $p$ is prime.

Solution 4:

There is a whole book about that ranging over various fields of mathematics. Mathematical Fallacies, Flaws and Flimflam by Edward J. Barbeau. Love it! And for all of the "proofs", it takes care of the first three of your bullets.