Where do I start with this?

#vwxyz=v+w+x+y+z#

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

2 Answers
George C.
Apr 4, 2018

Maximum for natural numbers is #x=5#

Explanation:

Given:

#vwxyz = 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 #vwxyz=1 < 5 =v+w+x+y+z#

  • If #v=w=x=y=z=2# then #vwxyz=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" "# 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:

#2x = x+5" "=>" "color(blue)(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:

#4x = x+6" "=>" "color(blue)(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:

#8x = x+7" "=>" "color(blue)(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^4x = x+4v" "=>" "x = (4v)/(v^4-1)#

Note that:

#lim_(v->1+) (4v)/(v^4-1) = +oo#

So #x# can be arbitrarily large.

George C.
Apr 4, 2018

Maximum value #x=5#

Explanation:

Another approach...

Given:

#vwxyz = v+w+x+y+z" "# 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" "# 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:

#((wyz)v - 1)x = v+(w+y+z)#

So:

#x = (v+(w+y+z))/((wyz)v-1)#

#color(white)(x) = ((wyz)v-1+1+(wyz)(w+y+z))/((wyz)((wyz)v-1))#

#color(white)(x) = 1/(wyz)+(1+(wyz)(w+y+z))/((wyz)((wyz)v-1))#

So:

#(del x)/(del v) = -(1+(wyz)(w+y+z))/((wyz)v-1)^2#

Note that if #v, w, y, z >= 1# with at least one #> 1# then the denominator is positive and #(del x)/(del 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...

#2x = x+5" "=>" "color(blue)(x=5)#

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