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
Explanation:
Solving
for
Regarding the solution for
regarding the solution for
so
or
and
For system
the feasible solutions are
then
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
Similarly, if we let
#q^2 = c#
#q^3 = d#
Then:
#2^6 = 64 = c - a = q^2-p^2 = (q-p)(q+p)#
Both
So the only possible factorings of
#{ ((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
#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))#