How to prove that $ab=a^2+4ab+4b^2$ is not true?

Question

Basically the title, this problem is from statistics so $a$ and $b$ are natural numbers including 0.

Answer 1

Hint

$$\implies a^2+3ab+4b^2=0$$

$$\iff0=4(a^2+3ab+4b^2)=(2a+3b)^2+7b^2$$

Answer 2

Well, the claim is that "$ab=a^2+4ab+4b^2$ for all natural numbers $a,b\in \mathbb{N}$." As pointed out in the comments, it suffices to provide a single pair of natural numbers for which this fails to show the statement does not hold. If $a=b=1$, then the claim states: $$ 1=1+4+4=9$$ which is a false statement.

Answer 3

Proof by counterexample. To help find values to select for this, rearrange the equation so that all variables are on one side: $$ ab=a^2+4ab+4b^2 $$ $$ 0=a^2+3ab+4b^2 $$ So, if $ a $ and $ b $ are both positive numbers, then $ a^2+3ab+4b^2 $ will be a positive number and successfully disprove the equation. For simplicity, let's choose $ a = 1 $ and $ b = 1 $ and substitute in these values into the original equation: $$ 1 \times 1=1^2+4\times 1 \times 1+4\times 1^2 $$ $$ 1 =1+4+4 $$ $$ 1 =9 $$ Since $ 1 \neq 9$, then $ a = 1 $ and $ b = 1 $ provide a sufficient counter example.

Answer 4

Note that the degree of every term is 2. Let $a=xb$ to replace a variable. Since $a$ and $b$ are integers, note that $x$ must be rational.

$$xb^2=(x^2+4x+4)b^2$$

Either $b=0$, and hence $a=0$, or

$$x=x^2+4x+4$$

$$0=x^2+3x+4$$

which is now a quadratic in one equation. One can then easily verify that this quadratic has no rational solutions by the rational roots theorem. Furthermore, since the discriminant is negative, there aren't any real solutions.

So the only real solution is $a=b=0$.