In the equation #w^3+x^3+y^3=z^3#, #w^3, x^3, y^3# and #z^3# are distinct, consecutive positive perfect cubes listed in ascending order. What is the smallest possible value of #z#?

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Answer 1 · 2017-02-18T21:34:42

#w=3,x=4,y=5# and #z=6#

Calling the first number as #n# we have

#n^3+(n+1)^3+(n+2)^3=m^3# or

#3(n^3+3n^2+5n+3)=m^3#

so we have #m = 3k# then

#n^3+3n^2+5n+3=3^2k^3# or

#(n^2+5)n = 3(3k^3-n^2-1)#.

At this point, the first guess is for #n=3# then

#3^2+5=3k^3-9-1# resulting in #k^3=8=2^3#

so #m = 3k =6# so the numbers are

#w=3,x=4,y=5# and #z=6#