Difference of Squares Cannot Equal 1

Problem Statement

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

Hints

Factor the left side as a difference of squares: $(x-y)(x+y) = 1$ .

For this product to equal $1$ , what must be true about the integer factors $(x-y)$ and $(x+y)$ ? Consider all possible integer factor pairs of $1$ .

Solutions

Factor the equation:

x^2 - y^2 = 1 \implies (x-y)(x+y) = 1

Since $x$ and $y$ are integers, both $(x-y)$ and $(x+y)$ are integers.

For the product of two integers to equal $1$ , the only possibilities are:

$x - y = 1$ and $x + y = 1$ , or
$x - y = -1$ and $x + y = -1$

Case 1: $x - y = 1$ and $x + y = 1$

Adding these equations: $2x = 2$ , so $x = 1$ .

Substituting back: $1 + y = 1$ , so $y = 0$ .

Since $y = 0$ is not a positive integer, this case fails.

Case 2: $x - y = -1$ and $x + y = -1$

Adding these equations: $2x = -2$ , so $x = -1$ .

Since $x = -1$ is not a positive integer, this case fails.

Conclusion: Neither case yields a solution with both $x$ and $y$ positive integers.

Therefore, there are no positive integers $x, y$ such that $x^2 - y^2 = 1$ . $\blacksquare$

Takeaways

Reconstruct the full proof from the hint and the solution outline, and justify every transformation explicitly.
Check edge cases and verify where each assumption is used in the argument.

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 Probability: https://vumaths.com/booklets/hsc-probability/
HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/

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