answer:Apply the difference of squares as used in the original problem: [ 17^4 + 2 times 17^2 + 1 - 16^4 = (17^2 + 1)^2 - (16^2)^2. ] Using the identity ( a^2 - b^2 = (a-b)(a+b) ): [ = (17^2 + 1 - 16^2)(17^2 + 1 + 16^2). ] Calculate each term: [ 17^2 = 289, quad 16^2 = 256, quad 17^2 + 1 = 290, quad 16^2 = 256. ] So, [ (17^2 + 1 - 16^2) = (290 - 256) = 34, quad (17^2 + 1 + 16^2) = (290 + 256) = 546. ] Thus, [ (17^4 + 2 times 17^2 + 1 - 16^4) = 34 times 546. ] Factoring 546: [ 546 = 2 times 273 = 2 times 3 times 91 = 2 times 3 times 7 times 13. ] The prime factors of the expression are 2, 3, 7, 13, and 17 (since 17 divides 34). The largest prime factor is (boxed{17}).

answer:To prove that for all ( n in mathbb{N} ), the number ( 2^{n+1} ) is in exactly ( n ) Pythagorean triples, we will analyze the properties of Pythagorean triples and the role of powers of 2 within them. 1. **Primitive Pythagorean Triples**: A primitive Pythagorean triple ((a, b, c)) is a set of three positive integers such that (a^2 + b^2 = c^2) and (gcd(a, b, c) = 1). Any Pythagorean triple can be derived from a primitive one by multiplying all elements by a common factor. 2. **Case Analysis**: We need to consider two cases for the position of (2^k) in the Pythagorean triple: - **Case 1: (2^k) is the largest member of the triple**: If (2^k) is the largest member, then we have: [ x^2 + y^2 = (2^k)^2 = 2^{2k} ] Taking this equation modulo 4, we get: [ x^2 + y^2 equiv 2^{2k} pmod{4} ] Since (x) and (y) are integers, (x^2) and (y^2) can only be 0 or 1 modulo 4. Therefore, the left-hand side (LHS) can be 0, 1, or 2 modulo 4, but the right-hand side (RHS) is 0 modulo 4. Hence, there are no solutions in this case. - **Case 2: (2^k) is one of the smaller members of the triple**: If (2^k) is one of the smaller members, then the other two members must be odd. We have: [ 2^{2k} + y^2 = z^2 implies 2^{2k} = z^2 - y^2 = (z - y)(z + y) ] Since (2^{2k}) is a power of 2, both (z - y) and (z + y) must also be powers of 2. Let (z - y = 2^a) and (z + y = 2^b) with (a < b). Then: [ z = frac{2^a + 2^b}{2} quad text{and} quad y = frac{2^b - 2^a}{2} ] For (z) and (y) to be integers, (a) and (b) must be such that (a) and (b) are both even or both odd. Let (a = k-1) and (b = k+1), then: [ z = 2^{k-1} + 2^{k-1} = 2^k quad text{and} quad y = 2^{k-1} - 2^{k-1} = 0 ] This is not possible since (y) must be positive. Therefore, we need to adjust our approach. Instead, consider: [ z = 2^{k-1} + 1 quad text{and} quad y = 2^{k-1} - 1 ] This gives us a valid Pythagorean triple ((2^k, 2^{k-1} - 1, 2^{k-1} + 1)). 3. **Counting the Triples**: For each (k) from 1 to (n), there is exactly one primitive Pythagorean triple containing (2^k). Therefore, for (2^{n+1}), there are exactly (n) such triples. (blacksquare)

answer:We start by recognizing the general form of a cubic binomial expansion (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3. Here, x=8 and y=a. This expression can be rewritten and compared to our problem: - (8+a)^3 = 8^3 + 3(8^2)a + 3(8)a^2 + a^3. - Coefficients in the problem are modified: 4a replaces 3a, and 6a^2 replaces 3a^2. **Set a=2 and substitute into our expression:** - Calculate each term: - 8^3 = 512 - 4a(8^2) = 4 cdot 2 cdot 64 = 512 - 6a^2 cdot 8 = 6 cdot 4 cdot 8 = 192 - a^3 = 2^3 = 8 - Sum these values: - 512 + 512 + 192 + 8 = 1224 Thus, the value of the expression 8^3 + 4(2)(8^2) + 6(2^2)(8) + 2^3 is boxed{1224}.

answer:If Ethan finished the race, he ran the entire 1 km, which is equivalent to 1000 meters. At that time, Arianna was 816 meters behind him. To find out how far Arianna was from the start line when Ethan finished, we subtract the distance she was behind from the total distance of the race. Distance Arianna ran = Total distance of the race - Distance Arianna was behind Ethan Distance Arianna ran = 1000 meters - 816 meters Distance Arianna ran = 184 meters So, Arianna was boxed{184} meters from the start line when Ethan finished the race.