# Where do I start with this?

## $v w x y z = v + w + x + y + z$ Find the maximum possible value of $x$ where $v , w , x , y , z > 0$

##### 2 Answers
Apr 4, 2018

Maximum for natural numbers is $x = 5$

#### Explanation:

Given:

$v w x y z = v + w + x + y + z$

For solutions in natural numbers, we can look at possible combinations of small values.

Note that:

• If $v = w = x = y = z = 1$ then $v w x y z = 1 < 5 = v + w + x + y + z$

• If $v = w = x = y = z = 2$ then $v w x y z = 32 > 10 = v + w + x + y + z$

The minimum possible values of $v , w , y , z$ are $v = w = y = z = 1$, resulting in:

$x = x + 4 \text{ }$ which has no solutions.

Considering the case where exactly one of $v , w , y , z > 1$, and putting $v = 2$, $w = y = z = 1$ we get:

$2 x = x + 5 \text{ "=>" } \textcolor{b l u e}{x = 5}$

Giving $v$ larger values either results in no integer solution or smaller integer solutions for $x$.

Considering the case where exactly two of $v , w , y , z > 1$, and putting $v = w = 2$, $y = z = 1$ we get:

$4 x = x + 6 \text{ "=>" } \textcolor{b l u e}{x = 2}$

Giving $v , w$ larger values results in smaller values of $x$ or no integer solution.

Considering the case where exactly three of $v , w , y > 1$, and putting $v = w = y = 2$, $z = 1$ we get:

$8 x = x + 7 \text{ "=>" } \textcolor{b l u e}{x = 1}$

Giving $v , w , y$ larger values results in no integer solution for $x$.

Footnote

If $v , w , x , y , z > 0$ are not restricted to integers, then simplify the problem by considering the case where $v = w = y = z$ and hence:

${v}^{4} x = x + 4 v \text{ "=>" } x = \frac{4 v}{{v}^{4} - 1}$

Note that:

${\lim}_{v \to 1 +} \frac{4 v}{{v}^{4} - 1} = + \infty$

So $x$ can be arbitrarily large.

Apr 4, 2018

Maximum value $x = 5$

#### Explanation:

Another approach...

Given:

$v w x y z = v + w + x + y + z \text{ }$ with $v , w , x , y , z > 0$

Looking for solutions in positive integers.

Note that if $v = w = y = z = 1$ then:

$x = x + 4 \text{ }$ which has no solutions.

So at least one of $v , w , x , y , z > 1$.

Considering the given equation as expressing $x$ as a function of $v$, with the other variables constant, we have:

$\left(\left(w y z\right) v - 1\right) x = v + \left(w + y + z\right)$

So:

$x = \frac{v + \left(w + y + z\right)}{\left(w y z\right) v - 1}$

$\textcolor{w h i t e}{x} = \frac{\left(w y z\right) v - 1 + 1 + \left(w y z\right) \left(w + y + z\right)}{\left(w y z\right) \left(\left(w y z\right) v - 1\right)}$

$\textcolor{w h i t e}{x} = \frac{1}{w y z} + \frac{1 + \left(w y z\right) \left(w + y + z\right)}{\left(w y z\right) \left(\left(w y z\right) v - 1\right)}$

So:

$\frac{\partial x}{\partial v} = - \frac{1 + \left(w y z\right) \left(w + y + z\right)}{\left(w y z\right) v - 1} ^ 2$

Note that if $v , w , y , z \ge 1$ with at least one $> 1$ then the denominator is positive and $\frac{\partial x}{\partial v} < 0$

So increasing the value of $v$ will decrease the value of $x$.

Similarly for $w , y , z$.

Hence the maximum value for $x$ is obtained with $v = 2$, $w = y = z = 1$ or permutations thereof...

$2 x = x + 5 \text{ "=>" } \textcolor{b l u e}{x = 5}$