Prove that #(5^(2n+1)+2^(2n+1))# is divisible by #7 \ \ AA n in NN#?

The proof is a little tricky, so I've typed something up below in case you would like a solution.

Proof. We will prove by induction that, #AA# #n>=0 in NN#,

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( 1 ) #color(white)(spacespacespace)7|(5^(2n+1)+2^(2n+1))#

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By the definition of "divides," #EE# #m in ZZ# s.t. #7m=5^(2k+1)+2^(2k+1)#.

Base case : When #n=1#, the right-hand side of ( 1 ) is #5^3+2^3=125+8=133#, and #7|133#. So, ( 1 ) is true for #n=1#.

Induction Step: Let #k in ZZ# be given and suppose ( 1 ) is true for #n=k#. Then

#7m=5^(2k+1)+2^(2k+1)#

#=>7m=5^(2(k+1)+1)+2^(2(k+1)+1)#

#=>7m=5^(2k+3)+2^(2k+3)#

#=>7m=5^2*5^(2k+1)+2^(2k+3)#

#=>7m=5^2(5^(2k+1)+2^(2k+1))-25(2^(2k+1))+4(2^(k+1))#

#=>7m=5^2(7m)-21(2^(2k+1))#

#=>7m=7(25m)-7(3*2^(2k+1))#

#=>7m=7(25m-3*2^(2k+1))#

By the closure property of integers, #(25m-3*2^(2k+1)) in ZZ#. Thus, ( 1 ) holds for #n=k+1# and the proof of the induction step is complete. #color(white)(spacespacespacesp)# ∎

Define #U_n# by;

# U_n = 5^(2n+1) + 2^(2n+1) #

Then our aim is to show that #U_n# is divisible by #7 AA n in NN#

We can prove this assertion by Mathematical Induction

When #n=0# the given result gives:

# U_n = 5^1 + 2^1 = 7 #

So the given result is true when #n=0#

Now, Let us assume that the given result is true when #n=k#, for some #k in NN#, in which case for this particular value of #k# we have:

# U_k = 5^(2k+1) + 2^(2k+1) #

And #U_k# is divisible by #7#, and so:

# 5^(2k+1) + 2^(2k+1)=7m#, say for some #m in NN#

# => 2^(2k+1) =7m-5^(2k+1)" " ..... (star)#

So, Using the definition of #U_n# we have:

# U_(k+1) = 5^(2(k+1)+1) + 2^(2(k+1)+1) #
# " " = 5^(2k+3) + 2^(2k+3) #
# " " = 5^(2k+3) + 2^(2k+1+2) #
# " " = 5^(2k+3) + 2^(2k+1)*2^2 #
# " " = 5^(2k+3) + 4*2^(2k+1) #
# " " = 5^(2k+3) + 4*(7m-5^(2k+1)) " " # (using #star#)
# " " = 5^(2k+3) + 4*7m-4*5^(2k+1) #
# " " = 5^(2k+1+2) + 4*7m-4*5^(2k+1) #
# " " = 5^(2k+1)*5^2 + 4*7m-4*5^(2k+1) #
# " " = 25*5^(2k+1) + 4*7m-4*5^(2k+1) #
# " " = (25-4)*5^(2k+1) + 4*7m #
# " " = 21*5^(2k+1) + 4*7m #
# " " = 7*3*5^(2k+1) + 4*7m #
# " " = 7(3*5^(2k+1) + 4m) #

Which is also clearly divisible by #7#

So, we have shown that if #U_n# is divisible by #7# for #n=k in NN#, then #U_n# is also divisible by #7# for #n=k+1#. But we initially showed that #U_n# was divisible by #7# for #n=0# and so it must also be true for #n=1, n=2, n=3, ... # and so on.

Hence, by the process of mathematical induction the given result is true for #n in NN# QED

A non inductive proof

#(a^(2n+1)+b^(2n+1))/(a+b) = a^(2n)-b a^(2n-1)+b^2a^(2n-2)+cdots+b^(2n)#

or

#(a^(2n+1)+b^(2n+1))=(a+b)(a^(2n)-b a^(2n-1)+b^2a^(2n-2)+cdots+b^(2n))#

Making #a=5# and #b=2# we conclude that

#(5^(2n+1)+2^(2n+1))equiv 0 mod 7#