Find the number of four tuples (a,b,c,d) of positive integers satisfying all three equations #a^3=b^2#, #c^3=d^2#, #c-a=64# ?

2 Answers
Jul 12, 2016

Answer:

#{a=15^2,b=15^3,c=17^2,d=17^3}# and
#{a=6^2,b=6^3,c=10^2,d=10^3}#

Explanation:

Solving

#{(a^3=b^2), (c^3=d^2), (c-a=64) :}#

for #b,c,d# we obtain the feasible (integer positive) solution

#b = a^(3/2), c = 64 + a, d = (64 + a)^(3/2)#

Regarding the solution for #b# then #a = m^2# and
regarding the solution for #d# then #64+a = n^2#

so

#n^2-m^2=64 = 2^6#

or

#(n+m)(n-m) = p cdot q = 2^6#

and

#n+m = p = {2^6,2^5,2^4}#

For system

#{ (n+m=p), (n-m=q) :}#

the feasible solutions are

#{n = 17, m = 15}# and #{n = 10,m = 6}#

then #a={15^2, 6^2}#
#d ={(sqrt (64+15^2))^3=17^3,(sqrt(64+6^2))^3=10^3}#
#c = {64+15^2=17^2,64+6^2=10^2}#
#b = {15^3, 6^3}#

Jul 13, 2016

Answer:

There are two, namely:

#color(blue)(""(15^2, 15^3, 17^2, 17^3)) = color(blue)(""(225, 3375, 289, 4913))#

#color(blue)(""(6^2, 6^3, 10^2, 10^3)) = color(blue)(""(36, 216, 100, 1000))#

Explanation:

Let:

#p = b/a#

Then:

#p^2 = b^2/a^2 = a^3/a^2=a#

#p^3 = b^3/a^3 = b^3/b^2 = b#

Then since #p# is a positive rational number whose square is a positive integer, #p# must be a positive integer.

Similarly, if we let #q = d/c#, then #q# is a positive integer with:

#q^2 = c#

#q^3 = d#

Then:

#2^6 = 64 = c - a = q^2-p^2 = (q-p)(q+p)#

Both #(q-p)# and #(q+p)# are positive integers and #(q-p) < (q+p)#.

So the only possible factorings of #2^6# give us:

#{ ((q-p) = 1), ((q+p) = 64) :}color(white)(XX)# hence #2q=65,color(white)(X) color(red)(cancel(color(black)(q=65/2)))#

#{ ((q-p) = 2), ((q+p) = 32) :}color(white)(XX)# hence #2q=34,color(white)(X) color(blue)(q = 17),color(white)(X) color(blue)(p = 15)#

#{ ((q-p) = 4), ((q+p) = 16) :}color(white)(XX)# hence #2q=20,color(white)(X) color(blue)(q = 10), color(white)(X) color(blue)(p = 6)#

So there are two possible tuples #(a,b,c,d)# satisfying the conditions, namely:

#color(blue)(""(15^2, 15^3, 17^2, 17^3)) = color(blue)(""(225, 3375, 289, 4913))#

#color(blue)(""(6^2, 6^3, 10^2, 10^3)) = color(blue)(""(36, 216, 100, 1000))#