Quadratic Residues Modulo 5

In number theory, quadratic residues tell us what remainders can appear when we square integers modulo a prime. This classic HSC problem asks you to find all possible quadratic residues modulo 5, then apply that insight to prove a divisibility statement about fourth powers. Working through it will strengthen your modular arithmetic, case analysis, and the crucial “hence” skill of building one result to prove another.

Problem Statement

• Given $a$ is integral and not divisible by 5, prove the remainder when $a^2$ is divided by 5 is either 1 or 4.
• Hence, given that $a, b$ are integral and not divisible by 5, prove that $a^4 - b^4$ is divisible by 5.

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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

Part (a): Proof by Cases

Since $a$ is not divisible by 5, we have $a \not\equiv 0 \pmod{5}$. To determine all possible remainders of $a^2$ when divided by 5, we can simply check the residues that $a$ itself can take modulo 5.

By the division algorithm, $a$ must be congruent to 1, 2, 3, or 4 modulo 5.

We check each case by squaring and reducing modulo 5:

Case 1: $a \equiv 1 \pmod{5}$

$$a^2 \equiv 1^2 \equiv 1 \pmod{5}$$

Case 2: $a \equiv 2 \pmod{5}$

$$a^2 \equiv 2^2 \equiv 4 \pmod{5}$$

Case 3: $a \equiv 3 \pmod{5}$

$$a^2 \equiv 3^2 \equiv 9 \equiv 4 \pmod{5}$$

Case 4: $a \equiv 4 \pmod{5}$

$$a^2 \equiv 4^2 \equiv 16 \equiv 1 \pmod{5}$$

In all cases, $a^2 \equiv 1$ or $4 \pmod{5}$.

Therefore, the remainder when $a^2$ is divided by 5 is either 1 or 4. $\\square$

Part (b): Using Part (a)

From part (a), for any integer $x$ not divisible by 5: $x^2 \equiv 1$ or $4 \pmod{5}$. We now need to find $a^4 = (a^2)^2$ modulo 5. Taking the result for $x=a$ and squaring it again gives us two possibilities:

• If $a^2 \equiv 1 \pmod{5}$, then $(a^2)^2 \equiv 1^2 \equiv 1 \pmod{5}$
• If $a^2 \equiv 4 \pmod{5}$, then $(a^2)^2 \equiv 4^2 \equiv 16 \equiv 1 \pmod{5}$

In either case we end up with the same residue. Thus $a^4 \equiv 1 \pmod{5}$ for any integer $a$ not divisible by 5.

Because $b$ satisfies the same condition, we immediately have $b^4 \equiv 1 \pmod{5}$.

Therefore:

$$\begin{aligned} a^4 - b^4 &\equiv 1 - 1 \pmod{5} \\ &\equiv 0 \pmod{5} \end{aligned}$$

Hence $5 \mid (a^4 - b^4)$. $\\square$

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Takeaways

• Systematic Case Analysis: For $5 \nmid a$, check all residues $a \equiv 1, 2, 3, 4 \pmod{5}$ systematically
• Squaring Congruences: If $a \equiv b \pmod{m}$, then $a^2 \equiv b^2 \pmod{m}$
• ``Hence'' Strategy: Part (b) builds directly on part (a)'s result; apply it twice to get $a^4 \equiv b^4 \equiv 1 \pmod{5}$
• Fermat's Little Theorem Preview: Result $a^4 \equiv 1 \pmod{5}$ for $\gcd(a,5)=1$ is special case of FLT

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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 Polynomials, HSC Sequences