"No integers $x$ and $y$ exist for $28x+7y=8$"

Question

What is the logical structure of this statement?

No integers $x$ and $y$ exist for $28x+7y=8.$

I'm not sure, but I think the answer is $$¬∃x\;∃y\;(x ∈ \mathbb Z ∧ y ∈ \mathbb Z ∧ 28x + 7y = 8).$$

Answer 1

Your answer (in blue) looks good. It is equivalent to Marius's suggested formalisation (in red; however, I don't think his comment's first sentence is correct):

\begin{align} &\color{blue}{¬\,∃x\;\;∃y\;\;(x ∈ \mathbb Z \quad∧\quad y ∈ \mathbb Z \quad∧\quad 28x + 7y = 8)} \\\equiv {}&∀x\;\;∀y\;\;¬\,(x ∈ \mathbb Z \quad∧\quad y ∈ \mathbb Z \quad∧\quad 28x + 7y = 8) \\\equiv {}&∀y\;\;(∀x\;\;(x \not∈ \mathbb Z \quad∨\quad (y \not∈ \mathbb Z \quad∨\quad 28x + 7y \not= 8))) \\\equiv {}&\color{red}{∀y\;\;(∀x\;\;(x ∈ \mathbb Z \implies (y ∈ \mathbb Z \implies 28x + 7y \not= 8)))} \\\equiv {}&∀y\;\;(∀x{∈} \mathbb Z \;\;(y ∈ \mathbb Z \implies 28x + 7y \not= 8)) \\\equiv {}&∀x{∈} \mathbb Z\;\; (∀y\;\;(y ∈ \mathbb Z \implies 28x + 7y \not= 8)) \\\equiv {}&\color\red{∀x{∈} \mathbb Z\;\; ∀y{∈} \mathbb Z \quad 28x + 7y \not= 8}\tag1 \\\equiv {}&\color{blue}{¬\,∃x{∈} \mathbb Z\;\; ∃y{∈} \mathbb Z \quad 28x + 7y = 8}\tag2 \\\equiv {}&\color{blue}{¬\,∃(x,y){∈} \mathbb Z^2 \quad 28x + 7y = 8}\tag3 \\\equiv {}&\color{red}{∀(x,y){∈}\mathbb Z^2 \quad 28x + 7y \not= 8}.\tag4 \end{align}

More plainly: $$28x + 7y = 8 \text{ has no integral solution set.}$$