An ingenuine use of Prime Number Theorem

The following problem appeared in Selection Test II of International Maths Olympiad Training Camp 2010, India :

Given an integer k > 1, show that there exists an integer n > 1 and distinct positive integers a₁, a₂ ,..., a_n ( > 1) such that the sums  ∑a_j and ∑φ(a_j) are both k-th powers of some integers. ( Here φ(m) denotes the number of positive integers less than m and relatively prime to m.)

The line of action I took was to select a_j 's which are square free so that φ(a_j) is related to a_j in a simple way [here if a_j = p₁ p₂ ... p_r  is the prime factorisation into distinct primes then φ(a_j) = (p₁-1)(p₂-1) ... (p_r -1) ]  .....
The next thought was to select a_j 's so that ∑φ(a_j) is also related to ∑a_j in a simple way so that "latter is a k-th power" easily implies "former is a k-th power" ....
Let  ∑a_j = λ^k
Then comes the main idea ..... why not take λ to be a number that can be written as a sum of 2 distinct primes in k different ways!! ...... write the above as .... ∑a_j = λ.λ....λ (k times) = (sum of primes in first way)(second way)......(kth way) .... expand the product of brackets and take each term in the expansion to be our a_j 's (each of which is square free and distinct)
And since in φ(a_j)'s prime factorisation is obtained by replacing each prime factor p of a_j by (p-1), we have ∑φ(a_j) = (λ-2)^k .

What remains to prove is a beautiful result that has emerged out :

Given k > 1, there exists a number λ that can be written as a sum of two distinct primes in at least k different ways.

The proof I present is a nice use of the Prime Number Theorem ( which states that lim_x→∞ ( π(x) log x ÷ x ) = 1 ; π(x) denotes the number of primes less than or equal to x). In fact the well known  elementary result π(n) > (n ÷ 6 log n)  [here n is an integer] is sufficient to prove it. The proof is due to Dr. M.A. Prasad who is a regular faculty at the IMOTC :

Proof:
There are  π(n) prime numbers less than or equal to n and thus C₂^π(n) ways of choosing two of them and summing them ... thus there are C₂^π(n) distinct sums less than or equal to 2n .... therefore at least          (C₂^π(n) ÷ 2n) of them are equal say equal to λ. What remains to show is that there is an n such that      (C₂^π(n) ÷ 2n) is at least k. But (C₂^π(n) ÷ 2n)  = [π(n)][π(n) – 1] ÷ 4n  >  (n – 6 log n) ÷ (12 log n)²  [ using π(n) > (n ÷ 6 log n) ] which is an increasing function and therefore (C₂^π(n) ÷ 2n) is an increasing function and thus is greater than k for large n.
We have thus obtained  λ that can be written as a sum of two distinct primes in at least k different ways.

Q.E.D.

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