Logical proposition for "Every positive integer can be written as the sum of 2 squares"

Why would you switch from "forall" to "exists" if you wanted to specify "$x$ is positive"?

You're going to want $$(\forall x > 0)(\exists a \exists b)(a^2+b^2 = x)$$ or, if your language doesn't let you formulate that, $$(\forall x)[x > 0 \to (\exists a \exists b)(a^2+b^2 = x)]$$

As an aside, the proposition is false: $3$ cannot be written as the sum of two squares. It is necessary and sufficient that primes $3 \pmod{4}$ appear only to even powers in the prime factorisation.

You can solve this question by using your previous answer,

$$P_{100}:= \exists\ x,y\ (x^2 + y^2 = 100)$$

where you replace $100$ by "any positive integer", let $z$:

$$\forall z>0\ P_z,$$ giving $$\forall z>0\ \exists\ x,y\ (x^2 + y^2 = z).$$

($x,y,z\in\mathbb Z$ is left implicit.)

You can formulate this statement as:

$\large\forall_{x\in\mathbb{N}}\exists_{a,b\in\mathbb{N}}:x=a^2+b^2$

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If you're worried about $0\in\mathbb{N}$ (which is a matter of definition), then you can use $\mathbb{Z^+}$ instead.

BTW, the statement itself is false.