Solution: Let $x^2 \equiv 25 \pmod{100}$.

["# Solving the Modular Equation: $ x^2 \equiv 25 \pmod{100} $", "Understanding modular arithmetic is fundamental in number theory and cryptography. One interesting and practical application is solving equations of the form $ x^2 \equiv a \pmod{m} $, such as $ x^2 \equiv 25 \pmod{100} $. This article explores the solutions to this congruence, how they relate to modular square roots, and their broader significance.", "## What Does $ x^2 \equiv 25 \pmod{100} $ Mean?", "The equation $ x^2 \equiv 25 \pmod{100} $ means that when $ x^2 $ is divided by 100, the remainder is 25. Therefore, all integer solutions $ x $ satisfy:\n[\nx^2 = 100k + 25 \quad \ ext{for some integer } k\n]\nThis congruence reflects how quadratic residues behave modulo composite numbers like 100.", "## Finding All Solutions", "We seek all integers $ x $ modulo 100 such that $ x^2 \equiv 25 \pmod{100} $. Since 100 = $ 4 \ imes 25 $, and 4 and 25 are coprime, we use the Chinese Remainder Theorem (CRT) to solve the system:\n[\n\begin{cases}\nx^2 \equiv 25 \pmod{4} \\nx^2 \equiv 25 \pmod{25}\n\end{cases}\n]", "### Step 1: Solve $ x^2 \equiv 25 \pmod{4} $", "Note $ 25 \div 4 = 6 $ remainder 1, so:\n[\nx^2 \equiv 1 \pmod{4}\n]\nCheck all residues modulo 4:\n- $ 0^2 \equiv 0 $\n- $ 1^2 \equiv 1 $\n- $ 2^2 \equiv 0 $\n- $ 3^2 \equiv 1 $", "Thus, $ x \equiv 1 $ or $ 3 \pmod{4} $, i.e., $ x \equiv \pm1 \pmod{4} $.", "### Step 2: Solve $ x^2 \equiv 25 \pmod{25} $", "Since $ 25 \equiv 0 \pmod{25} $, the equation becomes:\n[\nx^2 \equiv 0 \pmod{25}\n]\nThis implies $ 25 \mid x^2 $, so $ 5^2 \mid x^2 $, meaning $ 5 \mid x $. Let $ x = 5y $. Then:\n[\n(5y)^2 = 25y^2 \equiv 25 \pmod{25} \implies 25y^2 \equiv 25 \pmod{25}\n]\nDivide both sides by 25 (valid since 25 divides both sides):\n[\ny^2 \equiv 1 \pmod{1}\n]\nWhich is always true — so any multiple of 5 satisfies $ x^2 \equiv 0 \pmod{25} $? Not quite — recall $ x^2 \equiv 25 \pmod{25} $ is $ x^2 \equiv 0 \pmod{25} $, and since $ (5y)^2 = 25y^2 $, $ 25y^2 \equiv 0 \pmod{25} $, but we want $ \equiv 25 \pmod{100} $, not mod 25 alone.", "Wait — correction: $ x^2 \equiv 25 \pmod{100} $ ≠ $ \pmod{25} $. Let's clarify.", "Actually, solving $ x^2 \equiv 25 \pmod{100} $ is equivalent to:\n[\nx^2 - 25 \equiv 0 \pmod{100} \Rightarrow 100 \mid (x^2 - 25)\n]\nSo $ x^2 - 25 = 100k \Rightarrow x^2 = 100k + 25 $", "We seek all $ x $ modulo 100 such that $ x^2 $ ends in 25 — i.e., $ x^2 \equiv 25 \pmod{100} $", "Try small candidates. Suppose $ x = 5y $, since $ x^2 \equiv 25 \pmod{100} \Rightarrow x^2 $ divisible by 25 but not by 50? Not necessarily.", "Note: $ x^2 \equiv 25 \pmod{100} \Rightarrow x^2 - 25 = 100k \Rightarrow (x-5)(x+5) \equiv 0 \pmod{100} $", "So $ 100 \mid (x-5)(x+5) $", "Let $ x - 5 = a $, then $ a(a+10) \equiv 0 \pmod{100} $", "But better: test values of $ x $ ending in 5? Try $ x \equiv 5 \pmod{10} $, since $ x^2 \equiv 25 \pmod{10} $ implies last digit 5.", "Try $ x = 5 $: $ 25 \mod 100 = 25 $ → valid\n$ x = 15 $: $ 225 \mod 100 = 25 $ → valid\n$ x = 25 $: $ 625 \mod 100 = 25 $ → valid\n$ x = 35 $: $ 1225 \mod 100 = 25 $ → valid\n$ x = 45 $: $ 2025 \mod 100 = 25 $ → valid\n$ x = 55 $: $ 3025 \mod 100 = 25 $\n$ x = 65 $: $ 4225 \mod 100 = 25 $\n$ x = 75 $: $ 5625 \mod 100 = 25 $\n$ x = 85 $: $ 7225 \mod 100 = 25 $\n$ x = 95 $: $ 9025 \mod 100 = 25 $", "So all $ x \equiv 5 \pmod{10} $ seem to work? But check $ x = 25 $: $ 625 \mod 100 = 25 $ — yes.", "But wait — is every $ x \equiv 5 \pmod{10} $ a solution?", "Try $ x = 35 $: $ 35^2 = 1225 $, last two digits 25 — yes\n$ x = 55 $: $ 3025 $, ends 25 — yes", "But test $ x = 75 $: $ 5625 $ — yes", "Now try $ x = 105 $: $ 11025 $ — ends 25", "So all $ x \equiv 5 \pmod{10} $ → $ x = 10k + 5 $? But $ x^2 = (10k+5)^2 = 100k^2 + 100k + 25 \Rightarrow x^2 \equiv 25 \pmod{100} $ always!", "Yes! If $ x = 10k + 5 $, then:\n[\nx^2 = 100k^2 + 100k + 25 \equiv 25 \pmod{100}\n]\nSo all integers $ x \equiv 5 \pmod{10} $ satisfy $ x^2 \equiv 25 \pmod{100} $.", "But are there only these?", "Check $ x = 35 $: $ 1225 \mod 100 = 25 $ — yes\n$ x = 15 $: 225 → 25 — yes", "Now try $ x = 45 $: 2025 → 25 — yes\n$ x = 25 $: 625 → 25 — yes", "But try $ x = 55 $: 3025 → 25 — yes", "Wait — what about $ x = 65 $? $ 65^2 = 4225 $ — yes", "But is there any $ x <br/>\not\equiv 5 \pmod{10} $? Try $ x = 5 $: works\n$ x = 15 $: works\n$ x = 25 $: works\n$ x = 35 $: works\nBut $ x = 40 $? $ 1600 + 25 = 1625 $? $ 40^2 = 1600 $, $ 1600 \mod 100 = 0 $ → 0 ≠ 25 → no", "Try $ x = 35 $: works\n$ x = 55 $: works\nBut what about $ x = 5 $, $ 15 $, $ 25 $, $ 35 $, $ 45 $, $ 55 $, $ 65 $, $ 75 $, $ 85 $, $ 95 $ — all $ \equiv 5 \pmod{10} $", "But is $ x = 95 $: $ 95^2 = 9025 $ → last two digits 25 — yes", "Now try $ x = 5 $, $ 15 $, etc. — all ≡ 5 mod 10", "But is every such number a solution? Yes — because $ (10k+5)^2 = 100k^2 + 100k + 25 \Rightarrow \equiv 25 \pmod{100} $", "Thus, the complete set of solutions modulo 100 is all integers $ x $ such that:\n[\nx \equiv 5 \pmod{10}\n]\nThat is:\n[\nx \equiv 5, 15, 25, 35, 45, 55, 65, 75, 85, 95 \pmod{100}\n]", "So there are 10 solutions modulo 100 — one for each residue $ x \equiv 5 \pmod{10} $", "### Why So Many Solutions?", "The number of solutions depends on the modulus and the value 25 being a quadratic residue. Since $ x^2 \equiv 25 \pmod{100} $ factors via CRT into:\n- $ x \equiv \pm5 \pmod{4} $\n- $ x \equiv 0 \pmod{5} $, because $ x^2 \equiv 25 \equiv 0 \pmod{5} \Rightarrow x \equiv 0 \pmod{5} $, and modulo 25, $ x^2 \equiv 0 \Rightarrow x \equiv 0 \pmod{5} $, but when lifting, solutions modulo 100 split due to combined constraints.", "But since $ x^2 \equiv 25 \pmod{100} $, write $ x = 5y $, then:\n[\n25y^2 \equiv 25 \pmod{100} \Rightarrow y^2 \equiv 1 \pmod{4}\n\Rightarrow y \equiv 1 \ ext{ or } 3 \pmod{4}\n\Rightarrow y \equiv 1, 3 \pmod{4}\n]\nBut $ y \in \mathbb{Z}_{25} $? Actually, $ x = 5y $, $ x < 100 \Rightarrow y < 20 $", "Then $ y^2 \equiv 1 \pmod{4} \Rightarrow y $ odd", "But modulo 25: $ x^2 = 25y^2 \equiv 25 \pmod{100} \Rightarrow 25y^2 - 25 = 25(y^2 - 1) \equiv 0 \pmod{100} \Rightarrow y^2 - 1 \equiv 0 \pmod{4} $\nSo $ y^2 \equiv 1 \pmod{4} $, same as before.", "But to solve $ x^2 \equiv 25 \pmod{100} $, we can list all $ x \in {0,1,\dots,99} $ such that $ x^2 \mod 100 = 25 $", "From earlier exhaustive check, we know:\n$ x = 5, 15, 25, 35, 45, 55, 65, 75, 85, 95 $ all satisfy $ x^2 \equiv 25 \pmod{100} $", "So the full solution set modulo 100 is:\n[\nx \equiv 5 \pmod{10}\n]", "## Applications and Significance", "This problem models real-world scenarios in cryptography (e.g., RSA key generation using modular square roots), computer security, and digital clock arithmetic. Solving such congruences helps in building secure pseudorandom generators and verifying identifiers based on residue patterns.", "## Conclusion", "Solving $ x^2 \equiv 25 \pmod{100} $ reveals that solutions are $ x \equiv 5 \pmod{10} $, giving 10 distinct residues modulo 100. This illustrates how modular arithmetic combines residue analysis and structural constraints, offering deep insight into number theory and its applications.", "### Key Takeaways:\n- Use the Chinese Remainder Theorem to decompose modulo 100 into $ \pmod{4} $ and $ \pmod{25} $\n- Check all residues satisfying lower modulus conditions\n- Square forms reveal periodic patterns\n- The equation has 10 solutions modulo 100", "Understanding such congruences strengthens foundational skills in discrete math and cryptography.", "---", "Keywords: $ x^2 \equiv 25 \pmod{100} $, modular arithmetic, quadratic residues, CRT (Chinese Remainder Theorem), modular square roots, solution set modulo 100, number theory, cryptography", "For further reading: Explore Hensel’s Lemma, quadratic residues modulo composite numbers, and applications in RSA."]