No Positive Integer Difference of One

This problem explores a classic non‑existence proof using algebraic factorisation and integer properties—a skill that sits squarely inside the HSC Proofs topic. By turning the equation into a product of two integer factors, you’ll see how a seemingly open‑ended search reduces to checking just two simple cases. The technique you’ll gain here is one you can reuse whenever a Diophantine equation involves a difference of squares.

Problem Statement

Show that there are no positive integers $x$ and $y$ such that $x^2 - y^2 = 1$.

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Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

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Solutions

Proof by Factorization and Case Analysis

We start with the equation $x^2 - y^2 = 1$. This is a Diophantine equation—meaning we’re looking for integer solutions—so the natural first move is to rewrite the left‑hand side in a form that reveals the integer structure of the problem.

Step 1: Factor the left side

We factor $x^2 - y^2$ as a difference of squares:

$$x^2 - y^2 = (x-y)(x+y) = 1$$

Now we have a product of two integer expressions that must equal $1$.

Step 2: Analyze integer factor pairs

Since $x$ and $y$ are positive integers, both $(x-y)$ and $(x+y)$ are integers. We need their product to equal 1.

The only ways to write 1 as a product of integers are:

• $1 = 1 \times 1$
• $1 = (-1) \times (-1)$

Because $x$ and $y$ are positive, $x+y$ is always positive, but $x-y$ might be negative, so we must examine both sign possibilities.

Step 3: Case 1 - Both factors equal 1

If $x - y = 1$ and $x + y = 1$, adding these equations gives:

$$2x = 2 \implies x = 1$$

Subtracting: $2y = 0 \implies y = 0$

But $y = 0$ contradicts the requirement that $y$ is a positive integer.

Step 4: Case 2 - Both factors equal $-1$

If $x - y = -1$ and $x + y = -1$, adding these equations gives:

$$2x = -2 \implies x = -1$$

But $x = -1$ contradicts the requirement that $x$ is a positive integer.

Conclusion

All possible integer factorizations of 1 lead to violations of the positive integer requirement.

Therefore, there are no positive integers $x$ and $y$ satisfying $x^2 - y^2 = 1$.

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Takeaways

• Factorization Strategy: Recognize $x^2 - y^2 = (x-y)(x+y)$ as difference of squares
• Integer Factor Analysis: For product $(x-y)(x+y) = 1$, only factor pairs are $(1,1)$ and $(-1,-1)$
• Systematic Case Checking: Solve simultaneous equations $x-y = a$ and $x+y = b$ for each factor pair $(a,b)$
• Constraint Verification: Always check solutions against domain restrictions (here: positive integers)

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Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Vectors, HSC Inequalities