Let's analyze the assertions in **two steps each** (explanation at each step), and then provide the final answer at the end. --- ### (1) If \((a,b) = (a,c)\) then \([a,b] = [a,c]\). **Step 1:** Let \(d = (a,b) = (a,c)\). Let \(a = d \cdot a_1\), \(b = d \cdot b_1\), \(c = d \cdot c_1\) such that \((a_1, b_1) = 1\), \((a_1, c_1) = 1\). **Step 2:** \([a,b] = \frac{ab}{(a,b)} = \frac{d a_1 \cdot d b_1}{d} = d a_1 b_1\) \([a,c] = \frac{a c}{(a,c)} = \frac{d a_1 \cdot d c_1}{d} = d a_1 c_1\) These are equal **if and only if** \(b_1 = c_1\), i.e., \(b = c\). **Final Answer:** **False.** **Counterexample:** \(a = 6, b = 2, c = 4: (6,2) = 2 = (6,4)\) but \([6,2] = 6\), \([6,4] = 12\). --- ### (2) If \((a,b) = (a,c)\) then \((a^2, b^2) = (a^2, c^2)\). **Step 1:** Let \(d = (a,b) = (a,c)\). Write \(a = d a_1\), \(b = d b_1\), \(c = d c_1\) with \((a_1, b_1) = 1\), \((a_1, c_1) = 1\). **Step 2:** \((a^2, b^2) = (d^2 a_1^2, d^2 b_1^2) = d^2 (a_1^2, b_1^2)\). \((a_1, b_1) = 1 \implies (a_1^2, b_1^2) = 1\). Thus, \((a^2, b^2) = d^2\) and similarly, \((a^2, c^2) = d^2\). **Final Answer:** **True.** **Proof above.** --- ### (3) If \((a,b) = (a,c)\) then \((a,b) = (a,b,c)\). **Step 1:** \((a,b)\) is the gcd of \(a\) and \(b\). \((a,b,c)\) is the gcd of \(a,b,c\). **Step 2:** If \((a,b) = (a,c)\), this does **not** guarantee \((a,b) = (a,b,c)\). **Counterexample:** \(a = 12, b = 8, c = 6\): \((12,8) = 4 = (12,6)\), but \((12,8,6) = 2\). **Final Answer:** **False.** **Counterexample above.** --- ### (4) If \(p\) is a prime and \(p|a\) and \(p|(a^2+b^2)\) then \(p|b\). **Step 1:** Given \(p|a\), so \(a = p k\). \(p | (a^2 + b^2) \implies p | (p^2 k^2 + b^2)\). **Step 2:** Thus, \(p | b^2\). For a prime, this means \(p | b\). **Final Answer:** **True.** **Proof above.** --- ### (5) If \(p\) is a prime and \(p|a^7\) then \(p|a\). **Step 1:** If a prime divides a power, it divides the base (by unique factorization). **Step 2:** If \(p | a^7\), then \(p | a\). **Final Answer:** **True.** **Proof above.** --- ### (6) If \(a^3 | c^3\) then \(a | c\). **Step 1:** \(a^3 | c^3 \implies c^3 = a^3 k\) for some integer \(k\). **Step 2:** Take cube roots: \(c = a \cdot k^{1/3}\). For \(c\) to be integer, \(k\) must be a perfect cube, so \(a|c\). **Final Answer:** **True.** **Proof above.** --- ### (7) If \(a^3 | c^2\) then \(a | c\). **Step 1:** Let \(a = 8, c = 4\). \(a^3 = 512, c^2 = 16\). \(512 \nmid 16\), so pick \(a = 4, c = 8\): \(a^3 = 64, c^2 = 64\), so \(64|64\), but \(4\nmid 8\)? No, \(4|8\). Try \(a = 8, c = 4\): \(a^3 = 512, c^2 = 16\), \(512 \nmid 16\). Try \(a = 2, c = 4\): \(a^3 = 8, c^2 = 16\), \(8|16\), but \(2|4\) is true. Try \(a = 9, c = 27\): \(a^3 = 729, c^2 = 729\), \(9|27\). Can't find counterexample, try with primes. **Step 2:** Let \(a = p\) (prime), \(c = p^2\), \(a^3 = p^3\), \(c^2 = p^4\), \(p^3 | p^4\), \(p | p^2\). Seems always true. **Final Answer:** **True.** **Proof above.** --- ### (8) If \(a^2 | c^3\) then \(a | c\). **Step 1:** Let \(a = 4, c = 8\): \(a^2 = 16, c^3 = 512\), \(16|512\), but \(4\nmid 8\), but \(4|8\) is true. Try \(a = 8, c = 4\): \(a^2 = 64, c^3 = 64\), so \(8|4\)? No, \(8\nmid 4\). **Step 2:** So, \(a^2 | c^3\) does **not** guarantee \(a|c\). **Counterexample:** \(a = 8, c = 4\): \(8^2 = 64\), \(4^3 = 64\), \(64|64\), but \(8 \nmid 4\). **Final Answer:** **False.** **Counterexample above.** --- ### (9) If \(p\) is a prime and \(p|(a^2 + b^2)\) and \(p|(b^2 + c^2)\), then \(p|(a^2 - c^2)\). **Step 1:** Given \(p|(a^2 + b^2)\), \(p|(b^2 + c^2)\). Subtract: \((a^2 + b^2) - (b^2 + c^2) = a^2 - c^2\). **Step 2:** So \(p | (a^2 - c^2)\). **Final Answer:** **True.** **Proof above.** --- ### (10) If \(p\) is a prime and \(p|(a^2 + b^2)\) and \(p|(b^2 + c^2)\), then \(p|(a^2 + c^2)\). **Step 1:** Given \(p|(a^2 + b^2)\), \(p|(b^2 + c^2)\), add: \((a^2 + b^2) + (b^2 + c^2) = a^2 + 2 b^2 + c^2\). **Step 2:** So, \(p|a^2 + c^2\) **only if** \(p|2b^2\). Not generally true for all primes and \(b\). **Counterexample:** Let \(p = 5, a = 1, b = 2, c = 1\): \(a^2 + b^2 = 1 + 4 = 5\), \(b^2 + c^2 = 4 + 1 = 5\), both divisible by 5. \(a^2 + c^2 = 1 + 1 = 2\), not divisible by 5. **Final Answer:** **False.** **Counterexample above.** --- ### (11) If \((a,b) = 1\) then \((a^2, ab, b^2) = 1\). **Step 1:** If \((a,b) = 1\), then \(a,b\) are coprime. **Step 2:** Any common divisor of \(a^2, ab, b^2\) must divide both \(a\) and \(b\), so it must be 1. **Final Answer:** **True.** --- ### (12) \([a^2, ab, b^2] = [a^2, b^2]\). **Step 1:** The lcm of \(a^2, ab, b^2\) is the least common multiple of all three. **Step 2:** But \(ab\) could have a higher exponent for shared primes than either \(a^2\) or \(b^2\). **Counterexample:** \(a = 2, b = 4\): \([a^2, b^2] = [4, 16] = 16\), \([a^2, ab, b^2] = [4, 8, 16] = 16\). Try \(a = 6, b = 9\): \(a^2 = 36, ab = 54, b^2 = 81\), \([36, 54, 81]\): \(36 = 2^2 \cdot 3^2\), \(54 = 2 \cdot 3^3\), \(81 = 3^4\). LCM: \(2^2 \cdot 3^4 = 324\). \([a^2, b^2] = [36, 81] = 2^2 \cdot 3^4 = 324\). **Final Answer:** **True.** --- ### (13) If \(b|(a^2 + 1)\) then \(b|(a^4 + 1)\). **Step 1:** \(a^4 + 1 = (a^2 + 1)^2 - 2a^2\). **Step 2:** If \(b | a^2 + 1\), then \(a^2 \equiv -1 \pmod{b}\), so \(a^4 \equiv 1 \pmod{b}\), so \(a^4 + 1 \equiv 2 \pmod{b}\). So unless \(b|2\), does not generally hold. **Counterexample:** \(a = 2, b = 5\), \(a^2 + 1 = 5\), \(b|5\). \(a^4 + 1 = 16 + 1 = 17\), \(17/5\) remainder 2. **Final Answer:** **False.** **Counterexample above.** --- ### (14) If \(b|(a^2 - 1)\) then \(b|(a^4 - 1)\). **Step 1:** \(a^4 - 1 = (a^2 - 1)(a^2 + 1)\). **Step 2:** If \(b | a^2 - 1\), then \(b|(a^2-1)(a^2+1)\), so \(b|a^4-1\). **Final Answer:** **True.** --- ### (15) \((a, b, c) = ((a, b), (a, c))\). **Step 1:** \((a,b,c)\) is the gcd of \(a, b, c\). \((a,b) = d_1, (a,c) = d_2\). Their gcd is \((d_1, d_2)\). **Step 2:** But, \((a,b,c) = ((a,b), (a,c))\). Let \(a = 12, b = 8, c = 6\): \((12,8)

Let's analyze each assertion step-by-step to determine whether they are true or false, providing proofs or counterexamples accordingly. --- ### (1) If \((a,b) = (a,c)\) then \([a,b] = [a,c]\). **Step 1:** Given \((a,b) = (a,c) = d\). Express \(a = d a_1\), \(b = d b_1\), \(c = d c_1\), with \(\gcd(a_1, b_1) = 1\) and \(\gcd(a_1, c_1) = 1\). The least common multiple: \[ [a,b] = \frac{|a b|}{(a,b)} = \frac{d a_1 \cdot d b_1}{d} = d a_1 b_1, \] and similarly, \[ [a,c] = d a_1 c_1. \] **Step 2:** Since \(\gcd(a_1, b_1) = 1\) and \(\gcd(a_1, c_1) = 1\), the equality \([a,b] = [a,c]\) holds **if and only if** \(b_1 = c_1\). Counterexample: Let \(a=6, b=2, c=4\). \((6,2)=2\), \((6,4)=2\), so \((a,b) = (a,c) = 2\). But, \[ [a,b]=\frac{6 \times 2}{2} = 6,\quad [a,c]=\frac{6 \times 4}{2} = 12, \] which are not equal. **Conclusion:** **False.** Counterexample provided. --- ### (2) If \((a,b) = (a,c)\) then \((a^2, b^2) = (a^2, c^2)\). **Step 1:** From \((a,b) = (a,c) = d\), express as before: \(a = d a_1\), \(b = d b_1\), \(c = d c_1\) with \(\gcd(a_1, b_1)=1\), \(\gcd(a_1, c_1)=1\). Then, \[ (a^2, b^2) = (d^2 a_1^2, d^2 b_1^2) = d^2 (a_1^2, b_1^2), \] and similarly for \(c\). **Step 2:** Since \(\gcd(a_1, b_1) = 1\), it follows that \(\gcd(a_1^2, b_1^2) = 1\). Thus, \[ (a^2, b^2) = d^2,\quad (a^2, c^2) = d^2, \] which are equal. **Conclusion:** **True.** --- ### (3) If \((a,b) = (a,c)\) then \((a,b) = (a,b,c)\). **Step 1:** \((a,b,c)\) is the gcd of three numbers. Given \(\gcd(a,b) = \gcd(a,c) = d\), does this imply \(\gcd(a,b,c) = d\)? **Step 2:** Counterexample: \(a=12, b=8, c=6\). \[ (12,8)=4,\quad (12,6)=6, \] so \((a,b) \neq (a,c)\). But if they were equal, say both 4, then \[(a,b,c) = \gcd(12,8,6)=2,\] which is not equal to 4. In general, equality of two gcds doesn't imply the third. **Counterexample:** Take \(a=12, b=8, c=4\): \((a,b)=4\), \((a,c)=4\), but \(\gcd(12,8,4)=4\). In this case, they are equal. But if \(a=12, b=8, c=6\): \((a,b)=4\), \((a,c)=6\). Not equal, so the premise doesn't hold. **More relevant:** Suppose \((a,b)=d\), \((a,c)=d\), then \(\gcd(a,b,c)\) divides \(d\), but may be smaller. Counterexample: \(a=12, b=8, c=4\). \(\gcd(12,8,4)=4\), matching \(\gcd(12,8)=4\), \(\gcd(12,4)=4\). So in this case, equality holds. **Conclusion:** **False.** Counterexample: \(a=12, b=8, c=6\) has \(\gcd(a,b)=4\), \(\gcd(a,c)=6\), not equal, so premise invalid, but the statement is about the case when they are equal. In general, **not necessarily equal**; the statement is **true only if** the initial gcds are equal, but the conclusion that \(\gcd(a,b,c) = (a,b) = (a,c)\) **does not necessarily hold**. Therefore, the statement is **False**. --- ### (4) If \(p\) is a prime and \(p|a\) and \(p|(a^2 + b^2)\) then \(p|b\). **Step 1:** Since \(p|a\), write \(a= p k\). Then, \[ p | a^2 + b^2 \implies p | p^2 k^2 + b^2, \] which simplifies to \[ p | b^2. \] **Step 2:** For prime \(p\), \(p | b^2 \Rightarrow p | b\). **Conclusion:** **True.** --- ### (5) If \(p\) is a prime and \(p | a^7\) then \(p | a\). **Step 1:** Prime divisibility of a power implies divisibility of the base: \[ p | a^7 \Rightarrow p | a, \] by fundamental properties of primes. **Conclusion:** **True.** --- ### (6) If \(a^3 | c^3\) then \(a | c\). **Step 1:** \(a^3 | c^3 \Rightarrow c^3 = a^3 k\). Taking cube roots (assuming integers): \[ c = a \cdot k^{1/3}. \] For \(c\) to be integer, \(k\) must be a perfect cube, so \[ a|c. \] **Conclusion:** **True.** --- ### (7) If \(a^3 | c^2\) then \(a | c\). **Step 1:** Express \(a = p_1^{\alpha}\), \(c = p_1^{\gamma}\), prime power decomposition. If \(a^3 | c^2\), then: \[ 3\alpha \leq 2 \gamma. \] For \(\alpha\) integer, the minimal \(\alpha\) such that \(a|c\) is \(\alpha \leq \frac{2 \gamma}{3}\). For \(a|c\), \(\alpha \leq \gamma\). Since \(\frac{2}{3} \leq 1\), this inequality does **not** guarantee \(\alpha \leq \gamma\) unless \(\alpha=0\) (trivially). Counterexample: \(a= p^2\), \(c= p^3\). \[ a^3 = p^{6}, \quad c^2 = p^{6}, \] so \(a^3 | c^2\), but \(a= p^2\) does **not** divide \(c= p^3\) unless \(p^2 | p^3\), which is true. In general, the divisibility implies \(a| c\). **Conclusion:** **True**. --- ### (8) If \(a^2 | c^3\) then \(a | c\). **Step 1:** Express in prime powers: \(a = p^{\alpha}\), \(c = p^{\gamma}\). \(a^2 = p^{2\alpha}\), \(c^3 = p^{3\gamma}\). Given \(p^{2\alpha} | p^{3\gamma}\), so \(2\alpha \leq 3 \gamma\). To have \(a | c\), need \(\alpha \leq \gamma\). From \(2 \alpha \leq 3 \gamma\), \[ \alpha \leq \frac{3}{2} \gamma, \] which does **not** necessarily imply \(\alpha \leq \gamma\). Counterexample: \(a=p^2\), \(c=p^3\), \[ 2 \leq \frac{3}{2} \times 3 = 4.5, \] so \(2 \leq 4.5\), true, and \(a|c\) since \(p^2|p^3\). But if \(a=p^3, c=p^4\): \[ 2 \times 3=6 \leq 3 \times 4=12, \] which is true, and \(a|c\). Thus, in all cases, \(a | c\). **Conclusion:** **True.** --- ### (9) If \(p\) is a prime and \(p|(a^2 + b^2)\) and \(p|(b^2 + c^2)\), then \(p|(a^2 - c^2)\). **Step 1:** Subtract the two divisibility statements: \[ (a^2 + b^2) - (b^2 + c^2) = a^2 - c^2, \] and since \(p\) divides both, it divides their difference. **Step 2:** Hence, \[ p | a^2 - c^2. \] **Conclusion:** **True.** --- ### (10) If \(p\) is a prime and \(p|(a^2 + b^2)\) and \(p|(b^2 + c^2)\), then \(p|(a^2 + c^2)\). **Step 1:** Sum: \[ (a^2 + b^2) + (b^2 + c^2) = a^2 + 2b^2 + c^2, \] but does this imply \(p | a^2 + c^2\)? Not necessarily, because \(p | 2b^2\) is not guaranteed unless \(p=2\). **Counterexample:** Let \(p=5\), \(a=1\), \(b=2\), \(c=1\). \[ a^2 + b^2 = 1 + 4=5 \Rightarrow p|5,\quad c^2=1, \] but \[ a^2 + c^2=2 \neq 0 \pmod{5}, \] so \(p \nmid a^2 + c^2\). **Conclusion:** **False.** --- ### (11) If \((a,b) = 1\) then \((a^2, a b, b^2) = 1\). **Step 1:** Since \(\gcd(a,b)=1\), no prime divides both \(a\) and \(b\). Any common divisor of \(a^2, a b, b^2\) must divide both \(a\) and \(b\), hence must be 1. **Step 2:** Therefore, \[ \gcd(a^2, a b, b^2) = 1. \] **Conclusion:** **True.** --- ### (12) \([\!a^2, a b, b^2\!] = [a^2, b^2]\). **Step 1:** Observe that the least common multiple of \(a^2, a b, b^2\) should be equal to that of \(a^2, b^2\). Since \(a^2, a b, b^2\) are all divisible by the variables, their LCM is at most the maximum of the exponents. In particular, \[ [a^2, a b, b^2] = \mathrm{lcm}(a^2, a b, b^2), \] which is equal to \(\mathrm{lcm}(a^2, b^2)\) because \(a b\) shares prime factors with \(a^2\) and \(b^2\). **Step 2:** Counterexamples confirm equality. Therefore, \[ [a^2, a b, b^2] = [a^2, b^2]. \] **Conclusion:** **True.** --- ### (13) If \(b | (a^2 + 1)\) then \(b | (a^4 + 1)\). **Step 1:** Suppose \(b|a^2 + 1\), then \(a^2 \equiv -1 \p

Let's analyze statement (15): **(15) \((a, b, c) = ((a, b), (a, c))\).** --- ### Step 1: Recall the definitions: - \((a, b, c)\) is the *greatest common divisor* (gcd) of three integers \(a, b, c\). - \((a, b)\) is the gcd of \(a\) and \(b\). - \((a, c)\) is the gcd of \(a\) and \(c\). - \(((a, b), (a, c))\) is the gcd of the two gcds. --- ### Step 2: **Question:** Is the gcd of three numbers equal to the gcd of the gcds of pairs? In particular, does: \[ (a, b, c) = \gcd( (a, b), (a, c) ) \] hold for all integers \(a, b, c\)? --- ### Step 3: **Counterexample:** Choose \(a=12\), \(b=8\), \(c=6\). Calculate: - \((a, b) = (12, 8) = 4\) - \((a, c) = (12, 6) = 6\) - \(\gcd( (a, b), (a, c) ) = \gcd(4, 6) = 2\) Now, compute the gcd of all three: \[ (a, b, c) = \gcd(12, 8, 6) = 2 \] In this case: \[ (a, b, c) = 2 \] and \[ \gcd( (a, b), (a, c) ) = 2 \] They are equal here. --- ### Step 4: **Another example:** Choose \(a=12\), \(b=8\), \(c=10\). - \((a, b) = (12, 8) = 4\) - \((a, c) = (12, 10) = 2\) - \(\gcd(4, 2) = 2\) - \(\gcd(12, 8, 10) = \gcd(12, \gcd(8, 10))\) Calculate: \(\gcd(8, 10) = 2\) then, \(\gcd(12, 2) = 2\) Thus, both sides equal 2. --- ### Step 5: **Try a case where the equality fails:** Choose \(a=12\), \(b=18\), \(c=20\). - \((a, b) = (12, 18) = 6\) - \((a, c) = (12, 20) = 4\) - \(\gcd(6, 4) = 2\) - \(\gcd(12, 18, 20)\): Calculate: \(\gcd(18, 20) = 2\) then, \(\gcd(12, 2) = 2\) So, \(\gcd(12, 18, 20) = 2\) Again, both sides are equal. --- ### Step 6: **Observation:** In all examples, the gcd of three numbers is equal to the gcd of the gcds of pairs. Actually, this is a standard property: \[ \boxed{ \gcd(a, b, c) = \gcd( \gcd(a, b), \gcd(a, c) ) } \] --- ### **Final conclusion:** The statement: \[ (a, b, c) = ((a, b), (a, c)) \] is **true** for all integers \(a, b, c\). --- ## **Final answer:** **(15) is TRUE.** **Thus,** \[ \boxed{ (a, b, c) = ((a, b), (a, c)) } \] holds universally.