Quadratic Residues Modulo 5

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.

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

Part (a): Proof by Cases

Since $a$ is not divisible by 5, we have $a \not\equiv 0 \pmod{5}$ .

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

We check each case:

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. \hfill $\square$

Part (b): Using Part (a)

From part (a), for any integer $x$ not divisible by 5: $x^2 \equiv 1$ or $4 \pmod{5}$ .

Consider $a^4 = (a^2)^2$ :

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

Thus $a^4 \equiv 1 \pmod{5}$ for any integer $a$ not divisible by 5.

Similarly, $b^4 \equiv 1 \pmod{5}$ .

Therefore:

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

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

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

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 Polynomials: https://vumaths.com/booklets/hsc-polynomials/
HSC Sequences: https://vumaths.com/booklets/hsc-sequences/

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