No Positive Integer Difference of One

Problem Statement

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

Hints

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

Solutions

Proof by Factorization and Case Analysis

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

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:

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

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 $x^2 - y^2 = 1$ .

\hfill $\square$

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)

Further Readings

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

HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
HSC Vectors: https://vumaths.com/booklets/hsc-vectors/
HSC Inequalities: https://vumaths.com/booklets/hsc-inequalities/

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