answer:First, convert 62_6 to decimal: 62_6 = 6 cdot 6^1 + 2 cdot 6^0 = 36 + 2 = 38_{10}. Then, write the equation for 124_b in decimal form: 124_b = 1 cdot b^2 + 2 cdot b^1 + 4 cdot b^0 = b^2 + 2b + 4. Set these two equal since the problem states they are the same number: b^2 + 2b + 4 = 38. Solve for b: b^2 + 2b + 4 - 38 = 0 rightarrow b^2 + 2b - 34 = 0. Factorizing the quadratic, (b - 4)(b + 8.5) = 0. Thus, b = 4 or b = -8.5. We want the positive value, so b = boxed{4}.

answer:To solve this problem, let's first identify all the prime months in a year. The prime numbers that correspond to months are 2, 3, 5, 7, and 11. Therefore, the prime months are: - February (2) - March (3) - May (5) - July (7) - November (11) Next, we need to determine how many days in each of these months are prime dates. Remember, a prime date means both the day and the month are prime numbers. 1. **February (2)**: Since 2007 was not a leap year, February had 28 days. The prime numbers within this range are 2, 3, 5, 7, 11, 13, 17, 19, 23. Therefore, February had 9 prime dates. 2. **March (3), May (5), and July (7)**: These months have 31 days. The prime numbers within this range are the same as for February, with the addition of 29 and 31. Therefore, each of these months had 11 prime dates. 3. **November (11)**: This month has 30 days. The prime dates are the same as for February, plus 29. Therefore, November had 10 prime dates. Adding these together: - February: 9 prime dates - March, May, July (3 months): 11 times 3 = 33 prime dates - November: 10 prime dates Total prime dates in 2007: 9 + 33 + 10 = 52. Therefore, there were boxed{52} prime dates in total in 2007.

answer:Since we are given the recursive formula a_{n+1} = 4a_n + 3 with n in mathbb{N}^+, let's try to find a pattern by observing the next term after simplifying the given expression: [ begin{aligned} a_{n+1} &= 4a_n + 3 a_{n+1} + 1 &= 4(a_n + 1) end{aligned} ] Notice that if we add 1 to both sides, the right-hand side of the equation becomes 4(a_n + 1), which suggests that the sequence formed by (a_n + 1) is a geometric series with the first term being a_1 + 1 = 4 and a common ratio of 4. Therefore, the term of the new sequence is given by the formula for the n-th term of a geometric sequence: [ a_{n+1} + 1 = 4^n cdot (a_1 + 1) ] Since a_1 + 1 = 4, we get: [ a_{n+1} + 1 = 4^n cdot 4 ] So, simplifying and subtracting 1 from both sides, we get the general term for our original sequence {a_n}: [ a_{n+1} = 4^{n+1} - 1 ] Substituting n+1 with n, to find a_n, we have: [ a_n = boxed{4^n - 1} ]

answer:1. Suppose the quadratic equation (x^2 + ax + b + 1 = 0) has two positive integer roots (x_1) and (x_2). 2. Using Vieta's formulas, the sum and product of the roots related to the coefficients of the quadratic equation are given by: [ x_1 + x_2 = -a ] [ x_1 cdot x_2 = b + 1 ] 3. We need to show that (a^2 + b^2) is a composite number. According to Vieta's relations, we have: [ a = -(x_1 + x_2) ] [ b = x_1 cdot x_2 - 1 ] 4. Now, express (a^2 + b^2) in terms of (x_1) and (x_2): [ a^2 + b^2 = (x_1 + x_2)^2 + (x_1 x_2 - 1)^2 ] 5. Expanding both terms: [ (x_1 + x_2)^2 = x_1^2 + 2x_1 x_2 + x_2^2 ] [ (x_1 x_2 - 1)^2 = x_1^2 x_2^2 - 2 x_1 x_2 + 1 ] 6. Combine these expansions: [ a^2 + b^2 = x_1^2 + 2x_1 x_2 + x_2^2 + x_1^2 x_2^2 - 2x_1 x_2 + 1 ] 7. Simplify: [ a^2 + b^2 = x_1^2 x_2^2 + x_1^2 + x_2^2 + 1 ] 8. Recognize that: [ x_1^2 x_2^2 + x_1^2 + x_2^2 + 1 = left(x_1^2 + 1right) left(x_2^2 + 1right) ] 9. Notice ( left(x_1^2 + 1right) left(x_2^2 + 1right) ) is a product of two integers greater than 1 because (x_1) and (x_2) are positive integers. Both (x_1^2 + 1) and (x_2^2 + 1) are greater than 1, hence their product is composite. 10. Therefore, (a^2 + b^2) is a composite number. (blacksquare)