Difference of Squares Cannot Equal 1

Proving that an equation has no integer solutions is a fundamental skill in HSC Mathematics Extension 1 and 2, especially within the topic of Proof. In this problem, we'll see how factoring a difference of squares and analysing integer factor pairs leads to a clean, rigorous argument. By working through the proof, you'll strengthen your ability to handle contradictions and consider all possible cases.

Problem Statement

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

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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 $1$.

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Solutions

We start by recognising the left side as a difference of squares.

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. The equation states that the product of these two integer factors equals $1$. Multiplying two integers to give $1$ is very restrictive: the only way this can happen is if each factor is a divisor of $1$. Because $1$ has divisors $1$ and $-1$, the only possible integer factor pairs are $(1, 1)$ and $(-1, -1)$. Therefore, the only possibilities are:

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

Now we must examine each pair to see if it can produce positive $x$ and $y$.

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

To solve for $x$, we add the two equations, which eliminates $y$:

Adding the two equations eliminates $y$: $(x-y)+(x+y) = 1+1 \implies 2x = 2$, so $x = 1$.

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

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

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

Again adding the equations: $2x = -2$, so $x = -1$.

Substituting into $x+y=-1$ gives $-1 + y = -1$, so $y = 0$.

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

Conclusion: Neither possible factor pair yields a solution with both $x$ and $y$ positive integers. Hence, there are no positive integers $x, y$ such that $x^2 - y^2 = 1$. $\blacksquare$

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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.
• Recall that the only integer factor pairs of $1$ are $(1,1)$ and $(-1,-1)$; considering both positive and negative possibilities is crucial for complete integer proofs.

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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 Probability, HSC Mechanics