We find all \( x \in \{2,3,\dots,16\} \) such that \( x^4 \equiv 1 \pmod{17} \).

["Finding All Solutions to ( x^4 \equiv 1 \pmod{17} ) in the Range ( x = 2, 3, \dots, 16 )", "Here is a detailed exploration of all integers ( x ) from 2 to 16 such that\n[\nx^4 \equiv 1 \pmod{17}.\n]\nThis problem lies at the intersection of modular arithmetic and number theory, particularly stemming from Fermat’s Little Theorem and properties of multiplicative groups modulo a prime.", "---", "### Understanding the Problem", "We seek all integers ( x ) in the range ( 2 \leq x \leq 16 ) satisfying\n[\nx^4 \equiv 1 \pmod{17},\n]\nwith 17 being a prime number. Since 17 is prime, the multiplicative group modulo 17, denoted ( \mathbb{Z}<em 17="17">{17}^ ), consists of all nonzero residues ( {1, 2, \dots, 16} ) under multiplication modulo 17. By Fermat’s Little Theorem, every element satisfies ( x^{16} \equiv 1 \pmod{17} ) for ( x <br/>\not\equiv 0 \pmod{17} ).", "We are interested in solutions specifically to ( x^4 \equiv 1 \pmod{17} )—these are the fourth-order roots of unity modulo 17.", "---", "### Step 1: Use Group Theory Insight", "The multiplicative group ( \mathbb{Z}{17}^ ) is cyclic of order 16. The solutions to ( x^4 \equiv 1 \pmod{17} ) are precisely the elements of order dividing 4 in this group. That is, the cyclic subgroup ( { x \in \mathbb{Z}^ : x^4 \equiv 1 } ) has exactly ( \gcd(4, 16) = 4 ) elements?", "Wait — correction: in a cyclic group of order ( n ), the number of solutions to ( x^k \equiv 1 \pmod{p} ) is ( \gcd(k, n) ) only under special conditions, but actually it's that the subgroup has order ( n ), and the number of solutions is ( \gcd(k, n) ) if the group is cyclic — but more precisely, the number of solutions to ( x^k \equiv 1 ) is exactly ( \gcd(k, n) ) no, that’s incorrect.", "Actually, in a cyclic group of order ( n ), for any divisor ( d ) of ( n ), there are exactly ( d ) solutions to ( x^d \equiv 1 \pmod{p} ). So since ( \mathbb{Z}{17}^ ) is cyclic of order 16, the number of solutions to ( x^4 \equiv 1 \pmod{17} ) is ( \gcd(4, 16) = 4 )? No — wait, this is misleading.", "Clarification: In a cyclic group of order ( n ), the equation ( x^k = 1 ) has exactly ( \gcd(k, n) ) solutions? No — that is false. The correct fact is: in a cyclic group of order ( n ), the number of solutions to ( x^k = 1 ) is exactly ( \gcd(k, n) ). But wait — no: the number of solutions to ( x^k = 1 ) in a cyclic group of order ( n ) is ( \gcd(k, n) )? Actually, no.", "Correct Fact: In a cyclic group of order ( n ), for any ( k ), the number of solutions to ( x^k = 1 ) is exactly ( \gcd(k, n) ).\nBut more accurately: the group has a unique subgroup of order ( d ) for each ( d \mid n ), and within that subgroup, the equation ( x^k = 1 ) has ( \gcd(k, d) ) solutions. Translating: since the full group has order 16, the solutions to ( x^4 \equiv 1 \pmod{17} ) form a subgroup whose order divides 4 and is consistent with the group structure.", "But better: since ( \mathbb{Z}{17}^ \cong \mathbb{Z}{16} ), and we are solving ( x^4 \equiv 1 \pmod{17} ) for ( x = 1, 2, \dots, 16 ), we can simply test all residues or use efficient filtering.", "But there’s a smarter algebraic path.", "---", "### Step 2: Use Fermat’s Little Theorem and Group Structure", "Since 17 is prime, ( x^{16} \equiv 1 \pmod{17} ) for ( x <br/>\not\equiv 0 \pmod{17} ).\nWe want ( x^4 \equiv 1 \pmod{17} ), so ( x ) has order dividing 4 in the multiplicative group.", "Let ( g ) be a primitive root modulo 17. A known primitive root mod 17 is 3. So every nonzero residue mod 17 is a power of 3: ( x = 3^k ) for ( k = 0, 1, \dots, 15 ).", "Then\n[\nx^4 \equiv 1 \pmod{17} \iff (3^k)^4 = 3^{4k} \equiv 1 \pmod{17} \iff 16 \mid 4k \iff 4 \mid k.\n]\nSo ( k \equiv 0 \pmod{4} ).\nThus ( k = 0, 4, 8, 12 ), giving:\n[\nx = 3^0, 3^4, 3^8, 3^{12} \pmod{17}\n]", "Now compute these powers modulo 17:", "- ( 3^0 = 1 )\n- ( 3^1 = 3 ), ( 3^2 = 9 ), ( 3^3 = 27 \equiv 10 ), ( 3^4 = 3 \cdot 10 = 30 \equiv 13 \pmod{17} )\n- ( 3^8 = (3^4)^2 = 13^2 = 169 \equiv 169 - 10\cdot17 = 169 - 170 = -1 \equiv 16 \pmod{17} )\n- ( 3^{12} = 3^8 \cdot 3^4 = 16 \cdot 13 = 208 ). Now ( 208 \div 17 = 12\cdot17 = 204 ), so ( 208 - 204 = 4 ), thus ( 3^{12} \equiv 4 \pmod{17} )", "So the solutions are:\n[\nx \equiv 1, 13, 16, 4 \pmod{17}\n]", "These are exactly the fourth roots of unity in ( \mathbb{Z}_{17}^ ).", "---", "### Step 3: Verify All Solutions in Range ( x = 2 ) to ( 16 )", "We now list the solutions:\n- ( x = 1 ): excluded (we start from ( x = 2 ))\n- ( x = 4 ): included\n- ( x = 13 ): included\n- ( x = 16 ): included", "So the values of ( x \in {2, 3, \dots, 16} ) satisfying ( x^4 \equiv 1 \pmod{17} ) are:\n[\n\boxed{4,\ 13,\ 16}\n]", "Let’s verify each:", "- ( 4^4 = 256 ). ( 256 \div 17 = 15\cdot"]