For a positive integer n, let \tau(n) be the sum of its divisors (including 1 and itself), and let \phi(n) be the number of integers x, 1 \leq x \leq n, such that x and n are relatively prime. For example, if n=18, then \tau(18)=1+2+3+6+9+18=39 and \phi(18)=6 since the numbers 1,5,7,11,13, and 17 are relatively prime to 18.
(a) Prove that \phi(n) \tau(n)<n^{2} for every positive integer n>1.
(b) Determine all positive integers n such that \phi(n) \tau(n)+1=n^{2}.
(c) Prove that there are no positive integers n such that \phi(n) \tau(n)+2023=n^{2}.
Please note that the condition n>1 was mistakenly omitted in the original statement of the problem in part (a). Our markers awarded full credit for proving the statement for n>1 with or without any comments about the case n=1 for which we have \phi(n)=1 and \tau(n)=1 and therefore \phi(n) \tau(n)=1=n^{2}.