1979_Question_6

So to prove this claim, we will look at two different sequences of integers:
1+10^4+10^8+...+10^{4n} (the given sequence)
1+10^2+10^4+10^6+...+10^{2n}

Now
10^{4n}-1=(10^4-1)(1+10^4+10^8+...+10^{4n-4})
and
10^{2n}-1=(10^2-1)(1+10^2+10^4+...+10^{2n-2})

10^{4n}-1 factors as (10^{2n}+1)(10^{2n}-1) (difference of two squares)

So now we can write down an equation:
(10^4-1)(1+10^4+10^8+...+10^{4n-4})=(10^2-1)(1+10^2+10^4+...+10^{2n-2})(10^{2n}+1)

Dividing through by (10^2-1) we get:
101\cdot(1+10^4+10^8+...+10^{4n-4})=(1+10^2+10^4+...+10^{2n-2})(10^{2n}+1)
For n>2, if you divide a term on the RHS by 101, you will get a quotient that is greater than 1. Then (1+10^4+10^8+...+10^{4n-4}) can be written as the product of two factors so it is composite.

Now let’s prove that n=1 and n=2 don’t work.
n=1
This case does not work since n has to be at least 2 in the sequence.

n=2
101\cdot (1+10^4) = (1+10^2)(10^4+1)
10^4+1=73 \cdot 137 so it is also composite.

Thus all numbers in that sequence is composite and we are done.

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