Proving a theorem using its converse

Solution 1:

Say natural number $n$ is good if it can be writen as a sum of two squares.

Theorem: $n$ is good iff $2n$ is good.

Proof: If $n=a^2+b^2$ then $2n = (a+b)^2+(a-b)^2$ and we are done.

Now the converse. Say $2n$ is good. Then, by already proven part, also $4n$ is good, so $$4n = x^2+y^2$$ Since $x,y$ must be both even, we can write $a=x/2$ and $b=y/2$ and we are done.

Solution 2:

Let $f: A \to B$ be a function and $g: A \to A$, $h: B \to B$ be bijections.

Prove that $f$ is surjective if and only if $h \circ f \circ g$ is surjective.

Proof: Suppose $f$ is surjective .... we get $ h \circ f \circ g$ is surjective.

Now the converse: Suppose $h \circ f \circ g = :e$ is surjective. Since $g^{-1}$ and $h^{-1}$ are surjective then, by already proven part, we have $h^{-1} \circ e \circ g^{-1}=f$ is surjective.

Solution 3:

I've just found the following :

Let $K$ an infinite field. $f(x_0,...,x_n)\in K[x_0,...,x_n]$ is homogeneus of degree $d$ $\Longleftrightarrow$ $f(\lambda x_0,...,\lambda x_n) = \lambda^d f(x_0,...,x_n)$ for all $\lambda\in K$.

Proof: $\Rightarrow$) For each monomial of $f(\lambda x_0,...,\lambda x_n)$ you can take out a factor $\lambda^d$. Hence we have the statement (basically is the definition).

$\Leftarrow$) Suppose $f(x_0,...,x_n)=\sum_{i=1}^kf_{j_i}(x_0,...,x_n)$ where $f_{j_i}$ are homogeneus of degree $j_i$. Now we have: $$ f(\lambda x_0,...,\lambda x_n) = \sum_{i=1}^kf_{j_i}(\lambda x_0,...,\lambda x_n)\\ \lambda^d f(x_0,...,x_n) = \sum_{i=1}^k \lambda^{j_i}f_{j_i}(x_0,...,x_n) $$ where the operation in the LHS is the hypotesis and in the RHS we are using the arrow ($\Rightarrow$) of this proposition.

Hence the polynomial $t^d f(x_0,...,x_n) - \sum_{i=1}^k t^{j_i}f_{j_i}(x_0,...,x_n) \in K(x_0,...,x_n)[t]$ has infinite solutions ($K$ is infinite), so it is the $0$ polynomial.

Then in the RHS survives only the degree $d$ part and $f$ is homogeneus.