Repeated Square Roots

$a= \sqrt{b\sqrt{b\sqrt{b}}}$ . By squaring both sides repetedly, we get $a^8= b^7$ , or $b= a^{8/7}$ (note that b is a positive integer). Since b is a positive integer, a must be a perfect 7th power [Why must this be true? - Calvin].

But $a > 1$ . Since $a^k$ is monotonically increasing whenever k is a positive real, $a^{8/7}$ is a monotonically increasing function. The minimum value of b will be attained when the minimum value of a is attained. The minimum possible value of a is $2^7$ . If $a= 2^7, b= 2^8= 256$ . This is the minimum possible value of b.

The equation can be simplified as:

a^8 = b^7

=> b = a^(8/7)

And the minimum value of a greater than 1 must be 2^7.

So b = (2^7)^(8/7) = 2^8 = 256.

Note that $\begin{aligned} \sqrt{b\sqrt{b\sqrt{b}}}&=&\sqrt{b\sqrt{b\cdot b^{1/2}}}\\ &=&\sqrt{b\sqrt{b^{3/2}}}\\ &=&\sqrt{b\cdot b^{3/4}}\\ &=&\sqrt{b^{7/4}}\\ &=&b^{7/8}. \end{aligned}$ For $a$ to be an integer, $b$ must be a perfect $8^\text{th}$ power. Since $b\not=1$ , the minimum value of $b$ is $2^8=\boxed{256}$ .

Squaring the equation (since both sides are positive), $a^2=b\sqrt{b\sqrt{b}}$ Squaring again, $a^4=b^2\cdot b\sqrt{b}=b^3\sqrt{b}$ Squaring again, $a^8=b^6\cdot b=b^7$ Thus, $a$ is a 7th power, and $b$ is an 8th power. The smallest 8th power greater than 1 is $\boxed{256}$ .

R.H.S can be written as $b^{\frac{7}{8}}$ . Here 256 is the smallest number which has an 8th root > 1.

The expression is equivalent to $b^{\frac{1}{2}} \cdot (\sqrt{b\sqrt{b}})^{\frac{1}{2}}$ , which is equivalent to $b^{\frac{1}{2}} \cdot (b\sqrt{b})^{\frac{1}{4}}=b^{\frac{1}{2}} \cdot b^{\frac{1}{4}} \cdot b^{\frac{1}{8}}=b^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8}}=b^\frac{7}{8}$ . This means that the 8th root of b must be an integer. The smallest number that satisfies this is $2^{8}=256$

first we shall try to eliminate the surds and convert the expression into the integrals powers of $a$ and $b$

First square both sides, we get - $a^2 = b\sqrt(b\sqrt(b))$

Now again square both sides, we get $a^4 = b^3\sqrt(b)$

And finally again squaring third time, we get- $a^8=b^7$

So, $b = a^{8/7} = a\times a^{1/7}$

We know that $a$ and $b$ are integers $> 1$ .

To minimize $b$ , we need to minimize $a$ but we also know that $a$ should be a perfect $7^{th}$ power which is $128$ (because for $a^{1/7}=1\Rightarrow a=1$ which is not possible so for $a^{1/7}=2\Rightarrow a=128$

Hence, $b_{min}=128\times2= \boxed{256}$

$a$ $=$ $\sqrt {b\sqrt {b\sqrt {b}}}$

We know the property that $a \sqrt {b}$ $=$ $\sqrt {a^{2}b}$ ,therefore,

$a$ $=$ $\sqrt {b\sqrt {\sqrt b^{3}}}$

$a$ $=$ $\sqrt {\sqrt {b^{2}\sqrt b^{3}}}$

$a$ $=$ $\sqrt {\sqrt {\sqrt b^{7}}}$

$a^{2}$ $=$ $\sqrt {\sqrt {b^{7}}}$

$a^{4}$ $=$ $\sqrt {b^{7}}$

$a^{8}$ $=$ $b^{7}$

Now, to minimize the value of $b$ , we need to start off with numbers with the base of $2$ (e.g. $4$ , $16$ and so on). Since $LCM(7,8)$ $=$ $56$ , therefore,

$2^{56}$ $=$ $b^{7}$

$b$ $=$ $2^{8}$

$b$ $=$ $\boxed {256}$

First, simplify the right hand side. $a = \sqrt{b \sqrt{b \sqrt {b}}} = \sqrt{b \sqrt{b^\frac{3}{2}}} = \sqrt{b^\frac{7}{4}} = b^\frac{7}{8} = (b^\frac{1}{8})^7$ . Hence, for $a$ to be a positive integer, $b^\frac{1}{8}$ has to be a positive integer as well.

If $b^\frac{1}{8} = 1$ , then $b = 1^8 = 1$ , which doesn't satisfy the requirement that $b > 1$ . Therefore, the next smallest positive integer for $b^\frac{1}{8}$ is $2$ . The minimum value of $b$ can then be calculated to be $2^8 = \boxed{256}$ .

$\begin{aligned} \sqrt {b\sqrt{b^{\tfrac{3}{2}}}} &= \sqrt {b \cdot b^{\tfrac{3}{4}}} \\&= \sqrt {b^{\tfrac{7}{4}}} \\&= b^{\tfrac{7}{8}}. \end{aligned}$ Thus, $b$ must be an $8$ th power. The smallest such value of $b$ is $2^8 = \boxed {256}$ .

\sqrt{b} has to be a perfect square, \sqrt{b \cdot \sqrt{b}} has to be a perfect square and \sqrt{b \cdot \sqrt{b \cdot \sqrt{b}}} also has to be a perfect square.

Therefore, for \sqrt{b}, the b in the square root has to equal b^2 or b^4 or b^6 (or any other b^even number). However, that even number multiplied by another perfect square also has to be a perfect square. The second perfect square must also have b^even number. But that second perfect square times the first perfect square times another third perfect square must also be a perfect square.

Putting that in simpler terms, you need all three of those perfect squares to be b^even exponent . So, for the three numbers all to have an even exponent, as well as it to have the minimum value of b , you have to use the smallest even number: 2. Let's call the leftmost b as b 1, the middle *b* to be b 2 and the rightmost b to be b 3. If you want b 3, b 2, and b 1 all to be perfect squares, you have to start with the right-most number. If b 3 were equal to a b^2 (or 2^2 since we're using 2 now as the smallest positive even integer), then it would be 4. However, then b 2 would also be b^2 or 2^2 or 4, and 4 * 2 is equal to 2^3 or b^3 or 8, which isn't a perfect square. You'll need b 2 * b 3 to be equal to a perfect square, so that means they both have to have even exponents.

So, if you make b_3 equal to a 2^4 (since 4 is the next smallest positive even integer), then the square root of that is 2^2, or 4, and multiplying that by b^4 or 2^4 or 16 , makes 4 * 16 , or 64 (aka b^6 or 2^6 ), which also has an even exponent, thus making it a perfect square. The root of that however, is b^3 or 2^3 or 8 , not a perfect square.

Trying b^6 into b_3 won't work because the square root of that is b^3 , and multiplying b^3 * b^6 would get you b^9 , not a perfect square.

So finally, you try b^(b^3) or b^8 . That gets you 2^8 or 256 . The square root of 2^8 is b^4 or 2^4 or 16 , another perfect square. Multiplying b^4 by b_3 or b^8 gets you b^12 and that's another perfect square. The square root of b^12 would be b^6 and multiplying b^6 by b^8 again gets you b^14 , which is another perfect square, thus creating a positive integer for *a .

In conclusion, the main thing you need is for b 3, b 2, and b 1, as well as b 2 * \sqrt{b 3} and b 1 * \sqrt{b 2 * \sqrt{b 3}} to all be perfect squares, so that means that b_3 has to be at minimum b^8 . Using the smallest positive, even, integer 2, you can get b^8 to be 256, the answer to the problem.

This is a little long, but I believe it is mathematically sound! :)

From the given relation, we get a= b^(\frac {1}{2}) b^(\frac {1}{4}) b^(\frac{1}{8}). \Rightarrow a = b^(\frac{7}{8}). i.e.b = a^(\frac{8}{7}). For b to be integer, the minimum value of a must be 2^7 so that the powers cancel out which leaves us with b = 2^8 = 256. (Note: a,b >1 given)

Repeated squaring on both sides 3 times gives $a^8$ = $b^7$ This gives a^(8/7) = b , which means b should be 256

a= √(b√(b√b)) => a^8 = b^7

since a and b are greater than 1 => min value of a = 2^7 =128 and min value of b = 2^8 =256

a^8 = (2^7)^8 = 2^{56} b^7 = (2^8)^7 = 2^{56}

The equation can be simplified to a = b^(7/8) which yield to a^8 = b^7 [1]. Multiplying both sides by b, we have (b/a)^8 = b. Likewise, we have (b/a)^7 = a after multiplying both sides of [1] by 1/a. Hence, since a and b are both integers, b/a is an integer. Let b = ak, for some positive integer k. Then b = k^8 and a = k^7. Hence (a,b) is in the form (k^7, k^8), for some positive integer k. Hence the minimum value of b = k^8 > 1 is 2^8 = 256.

a = \sqrt{b\sqrt{b\sqrt{b}}} square both sides of equation 3 times to remove radical sign

a^{8} = b^{7} get the 7th root of both sides of equation

\sqrt[7]{a^{8}} = b a \times \sqrt[7]{a} = b

since the minimum integer is needed, let \sqrt[7]{a} = 2 then, you will get a = 128 128 \times 2 = b therefore, 256 = b

\sqrt{b* \sqrt{b* \sqrt{b}}}= \sqrt{b* \sqrt{b b^{1/2}}}= \sqrt{b \sqrt{b^{3/2}}}= \sqrt{b*b^{3/4}}= \sqrt{b^{7/4}}=b^{7/8} So for this to work out, a and b have to be integers. The smallest number that has an eighth root is 256 (which is 2). 2^7 = 128.

$\sqrt{b\sqrt{b\sqrt{b}}} = b^{7/8} \implies b^8=x \wedge x>1 \wedge x \in \mathbb{N} \implies x=2 \wedge b=256$

$a = \sqrt{b\sqrt{b\sqrt{b}}}$

Integer exponents are easier to deal with than square roots. We could square and divide by $b$ three times to make the above equation more manageable. Equivalently, we can raise both sides to the eighth power as follows:

$a^8 = \left(\sqrt{b\sqrt{b\sqrt{b}}}\right)^{8}$

$a^8 = \left(b\sqrt{b\sqrt{b}}\right)^{4}$

$a^8 = b^4 \left(b\sqrt{b}\right)^2$

$a^8 = b^4 \cdot b^2 \cdot b$

$a^8 = b^7$

$a^8$ is a perfect seventh power, so $a$ must be as well. Similarly, $b$ must be a perfect eighth power. $a$ and $b$ are both integers greater than $1$ . We want the smallest $b$ that is an eighth power greater than $1$ . This is

$2^{8} = \boxed{256}$

As we see that a^8=b^7=k(say) we are to find a number which has a power of the form 56n (n is any natural number other than one as said in the Question ) the lowest possible is 2^56 and b^7=2^56 (for smallest b)

Thus b=256

If $a$ is an integer greater than $1$ , $\sqrt{b}$ is also an integer.Therefore,we can say that $\sqrt{b}=x,b=x^2$ .

$b\sqrt{b}$ is a square number too,since $\sqrt{b}=x,b$ must be $xy^2$ ,so that $b\sqrt{b}=x^2y^2,\sqrt{b\sqrt{b}}=xy.$

Next,use the same way. $b\sqrt{b\sqrt{b}}$ is also a square number,so $b=xyz^2$ ,and $b\sqrt{b\sqrt{b}}=(xyz)^2,$ and $\sqrt{b\sqrt{b\sqrt{b}}}=xyz$ .

Easier to say, $b=x^2=xy^2=xyz^2$ .

From the equation above,we can know that $x=y^2,y=z^2$ .

$b=z^{2\times2\times2}=z^8$

$z>1$ ,so the minimum value of $z$ is $2$ .

Minimum value of $b=2^8=\boxed{256}$

Simplification yields $a^8 = b^7$ . Thus, for some integer $k$ , $a=k^7$ and $b=k^8$ so that $k^{56} = k^{56}$ . As $a \neq 1$ , $k \geq 2$ , so let us say that $k=2$ as we are looking for the minimum integer value of $b$ . Then $b = k^8 = 2^8 = \boxed{256}$ .

$a=\sqrt{b\sqrt{b\sqrt{b}}}\Longrightarrow a=b^{7/8}\Longrightarrow b=a^{8/7}=(\sqrt[7]a)^8$

To let $b$ be a positive integer, $a$ must be a $7$ th power... So, $a = k^7$ for some positive integer $k$

To minimize $b$ , we must minimize $a$ , and to do that, we have to minimize $k$

$k = 1$ yields $a = 1$ , but the question says that $a>1$ and $a\in N$

Hence, $k=2$ ... So, $a = 2^7 =128$

Therefore, $b =\left(\sqrt[7]{128}\right)^8 = \left(\sqrt[7]{2^7}\right)^8 = 2^8 = \fbox{256}$

Let, $b=x^p$

$a = \sqrt{x^p \sqrt{x^p \sqrt{x^p} } }$

$\implies a = \sqrt{x^p \sqrt{x^p \times x^{ \frac{p}{2} } } }$

$\implies a = \sqrt x^p \sqrt{x^{ \frac{3p}{2} } }$

$\implies a = \sqrt x^p \times x^{ \frac{3p}{4} }$

$\implies a = \sqrt x^{ \frac{7p}{4} }$

$\implies a = x^{ \frac{7p}{8} }$

let $p = \boxed{8}$

$\implies a = x^7$ [ $x^7$ is an integer value]

Then $b=x^8$

And, $b>1$ ; so, $x >1$

Let $x=2$

And, minimum value of $b=2^8 = \boxed{256}$

In order to get rid of all of the square roots, we can raise both sides of the equation to the 8th power. To make computations more easier, we can use the equation $((a^2)^2)^2=\left( \left( \left(\sqrt{b\sqrt{b\sqrt{b}}} \right)^2 \right)^2 \right)^2$ . The first square will give us $(a^2)^2)^2=((b\sqrt{b\sqrt{b}})^2)^2$ . The next one further simplifies it to $(a^4)^2=(b^3\sqrt{b})^2$ . Doing this once more finally gives us $a^8=b^7$ . Since the $\gcd (8,7)=1$ , $a^8=b^7=x^{56}$ , where $x$ is some positive integer greater than $1$ . The smallest possible value that meets this restriction is when $x=2$ , so therefore $a^8=b^7=2^{56}$ . We want to find [the minimum value of] $b$ , so $b=2^8=\boxed{256}$ .