Problem :- Prove that there does not exist any integer n < 2310 such that n(2310-n) is a multiple of 2310.

I had written an alternative solution of this problem in the exam. Please go through the solution once and have your comments. Please inform about some improvements if necessary.

My solution :- We know that 2310 = 2*3*5*7*11.
So every prime factor in the prime factorization of 2310 occurs only once.
So , suppose that n(2310-n) is divisible by 2310.
Let p be a prime factor of 2310. So either p divides n or p divides 2310-n.
Let us divide the problem into two cases :-
Case 1:- p divides n :-
If p divides n, then as p divides 2310, then p divides 2310-n.
So n = pa and
2310-n = pb
So n(2310-n) = (p^2)ab
Hence (p^2) divides (2310-n)n
Case 2 :- p divides 2310-n
So, p must also divide n. Hence, in the same way, (p^2) again divides n(2310-n).

So we may conclude from both cases that if p divides 2310, then n(2310-n) is divisible by (p^2)
So if 2310=a*b*c*d.....z
where a......z are prime factors of 2310,
then n(2310-n) = (a^2)*(b^2)......(z^2)> (a*b*c....z)^2 > (2310)^2
But we know that,
n<2310
and 2310-n < 2310
So n(2310-n) < (2310)^2
So this leads to contradiction.
Q.E.D.

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TopNewestI think you don't have to divide the problem into two cases since the two cases are the same.

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