answer:1. **Restate the problem and definitions:** We need to find all integers ( m ) such that the inequality [ frac{phi(n,m)}{n} geq frac{phi(m)}{m} ] holds for every positive integer ( n ). Here, (phi(n,m)) denotes the number of positive integers less than or equal to ( n ) that are coprime with ( m ), and (phi(m)) is Euler’s totient function. 2. **Case 1: ( m ) is a power of a prime:** Let ( m = p^a ) where ( p ) is a prime and ( a geq 1 ). We need to show that the inequality holds for every ( n ). - For ( m = p^a ), the function (phi(m)) is given by: [ phi(p^a) = p^a left(1 - frac{1}{p}right) = p^a - p^{a-1} ] - For any ( n ), (phi(n, p^a)) counts the numbers up to ( n ) that are not divisible by ( p ). This is: [ phi(n, p^a) = n - leftlfloor frac{n}{p} rightrfloor ] - Therefore, we have: [ frac{phi(n, p^a)}{n} = frac{n - leftlfloor frac{n}{p} rightrfloor}{n} ] - Since (leftlfloor frac{n}{p} rightrfloor leq frac{n}{p}), it follows that: [ frac{phi(n, p^a)}{n} geq frac{n - frac{n}{p}}{n} = 1 - frac{1}{p} ] - We know that: [ frac{phi(p^a)}{p^a} = 1 - frac{1}{p} ] - Thus: [ frac{phi(n, p^a)}{n} geq frac{phi(p^a)}{p^a} ] - Hence, the inequality holds for ( m = p^a ). 3. **Case 2: ( m ) has more than one distinct prime factor:** Suppose ( m ) has more than one distinct prime factor, say ( m = p_1 p_2 cdots p_r ) with ( r > 1 ). - Consider ( n = p_1 p_2 cdots p_r - p_1 ). We need to evaluate (phi(n, m)) and compare it with (phi(m)). - The number of integers up to ( n ) that are coprime with ( m ) is: [ phi(n, m) = phi(n, p_1 p_2 cdots p_r) = phi(p_1 p_2 cdots p_r) - (p_1 - 1) ] because ( p_1 p_2 cdots p_r - i ) for ( 0 < i < p_1 ) are all relatively prime to ( m ). - We have: [ phi(p_1 p_2 cdots p_r) = p_1 p_2 cdots p_r left(1 - frac{1}{p_1}right)left(1 - frac{1}{p_2}right) cdots left(1 - frac{1}{p_r}right) ] - Therefore: [ frac{phi(m)}{m} = frac{phi(p_1 p_2 cdots p_r)}{p_1 p_2 cdots p_r} ] - We need to show: [ frac{phi(n, m)}{n} = frac{phi(p_1 p_2 cdots p_r) - (p_1 - 1)}{p_1 p_2 cdots p_r - p_1} < frac{phi(p_1 p_2 cdots p_r)}{p_1 p_2 cdots p_r} ] - This inequality simplifies to: [ frac{phi(p_1 p_2 cdots p_r) - (p_1 - 1)}{p_1 p_2 cdots p_r - p_1} < frac{phi(p_1 p_2 cdots p_r)}{p_1 p_2 cdots p_r} ] - Which is equivalent to: [ -(p_1 - 1)p_1 p_2 cdots p_r < -p_1 phi(p_1 p_2 cdots p_r) ] - Simplifying further: [ p_2 cdots p_r > (p_2 - 1) cdots (p_r - 1) ] - This inequality is clearly true since each ( p_i > 1 ). 4. **Conclusion:** The inequality holds for every positive integer ( n ) if and only if ( m ) is a power of a prime. The final answer is ( boxed{ m } ) is a power of a prime.

answer:We start by eliminating the square roots by isolating them and squaring both sides of the equation: [ frac{sqrt{8y}}{sqrt{6(y-2)}} = 3 ] Squaring both sides gives: [ left(sqrt{8y}right)^2 = left(3 sqrt{6(y-2)}right)^2 ] [ 8y = 54(y-2) ] Expanding the right side: [ 8y = 54y - 108 ] Solving for y: [ 54y - 8y = 108 ] [ 46y = 108 ] [ y = frac{108}{46} = frac{54}{23} ] Thus, y = boxed{frac{54}{23}}.

answer:1. Let's consider that ( n ) is a positive integer, and ( d ) is a positive divisor of ( 2n^2 ). 2. By definition of divisors, there exists a positive integer ( k ) such that: [ 2n^2 = k d ] 3. Suppose for contradiction that ( n^2 + d ) is a perfect square. That is, there exists an integer ( x ) such that: [ n^2 + d = x^2 ] 4. Substitute ( d ) using ( d = frac{2n^2}{k} ): [ n^2 + frac{2n^2}{k} = x^2 ] 5. Combining the terms gives: [ x^2 = n^2 + frac{2n^2}{k} ] [ x^2 = n^2 left(1 + frac{2}{k}right) ] Therefore, we can write: [ x^2 = n^2 cdot frac{k+2}{k} ] 6. Multiply both sides by ( k^2 ): [ k^2 x^2 = n^2 (k+2) k ] [ k^2 x^2 = n^2 (k^2 + 2k) ] 7. Looking at this equation mod ( p ) for any prime ( p ): [ p(k^2 x^2) = p(n^2 (k^2 + 2k)) ] So: [ 2p(kx) = 2p(n) + p(k^2 + 2k) ] 8. Since ( n ) and ( x ) are integers, their product with ( k ) will also be integers, which implies: [ p(k^2 + 2k) ] must be even. 9. Therefore, ( k^2 + 2k ) must be a sum of squares: [ k^2 + 2k = m^2 ] for some integer ( m ). 10. Notice that: [ k^2 < k^2 + 2k < (k+1)^2 ] This implies: [ k^2 < m^2 < (k+1)^2 ] 11. If ( k^2 + 2k ) would be a perfect square, ( m ) would have to be between ( k ) and ( k+1 ), but there are no integers between ( k ) and ( k+1 ). 12. Therefore, we conclude that ( k^2 + 2k ) is not a perfect square, leading to a contradiction in our assumption that ( n^2 + d ) is a perfect square. Conclusion: [ boxed{text{ntextsuperscript{2} + d is not a perfect square}} ]

answer:1. Let's start by examining the expression ((10^{4n^2 + 8} + 1)^2). We can rewrite it as: [ (10^X + 1)^2 quad text{where} quad X = 4n^2 + 8 ] 2. Expanding the square, we get: [ (10^X + 1)^2 = 10^{2X} + 2 cdot 10^X + 1 ] 3. Substituting (X = 4n^2 + 8), we have: [ 10^{2(4n^2 + 8)} + 2 cdot 10^{4n^2 + 8} + 1 ] 4. Notice that (10^{2(4n^2 + 8)}) is a 1 followed by (2(4n^2 + 8)) zeros, (2 cdot 10^{4n^2 + 8}) is a 2 followed by (4n^2 + 8) zeros, and the final 1 is just 1. When we add these together, the result is: [ 1underbrace{00ldots0}_{4n^2 + 8 text{ zeros}}2underbrace{00ldots0}_{4n^2 + 8 text{ zeros}}1 ] 5. The sum of the digits of this number is: [ 1 + 2 + 1 = 4 ] 6. Therefore, the sum of the digits in base ten of ((10^{4n^2 + 8} + 1)^2) is 4. The final answer is (boxed{4})