How will you show ? #5^n+1# is always divisible by 7 when n is a positive odd integer divisible by 3.

1 Answer
May 31, 2016

Answer:

(see below for proof)

Explanation:

If #n# is a positive odd integer divisible by #3#
we can replace #n# with #3(2p+1)# for some #"odd "p inZZ#

#5^n+1#
#color(white)("XXX")= 5^(3(p))+1#

#color(white)("XXX")=125^(p)+1#

Noting that #125=7xx18-1#

we have
#5^n+1#
#color(white)("XXX")=(7xx18-1)^(p)+1#

#rArr(5^n+1)_("mod "7)= (-1)^(p)+1#

but #(-1)^("any odd integer") = -1#

#rArr (5^n+1)_("mod "7)=-1+1=0#

#rArr (5^n+1)# is divisible by #7#