Show 2018th term of Sylvester's Sequence isn't a perfect square

Jan 2009
435
125
A sequence \(\displaystyle T_1, T_2, T_3 ... \) is defined by:
\(\displaystyle T_1 = 1\)
\(\displaystyle T_2 = 2\)
\(\displaystyle T_{n+1} = 1 + \prod_{i=1}^{n} T_{i}\) for all integers \(\displaystyle n \geq 2\)

Prove that \(\displaystyle T_{2018}\) is not a perfect square.

For context, this was given to me by a student, and is the last part of a four part question. The previous parts were:

(a) What is the value of \(\displaystyle T_5\)? (Ans: 1 + 1*2*3*7 = 43)
(b) Prove that \(\displaystyle T_{n+1} = T_n^2 - T_n + 1\) for \(\displaystyle n \geq 2\)
(Ans:
Observe that \(\displaystyle T_n^2 - T_n = T_n(T_n - 1) = T_n(\)\(\displaystyle \prod_{i=1}^{n-1} T_{i}) = T_{n+1} - 1\)
then add one on both sides)

(c) Prove that \(\displaystyle T_n + T_{n+1}\) is a factor of \(\displaystyle T_nT_{n+1} - 1\) for all integers \(\displaystyle n \geq 2\)
(Ans:
\(\displaystyle T_nT_{n+1} - 1 = T_n^3 - T_n^2 + T_n - 1 = (T_n^2 + 1)(T_n - 1) = (T_n + T_{n+1})(T_n - 1)\)

On part (d), which is to prove that \(\displaystyle T_{2018}\) is not a perfect square, I'm stumped. How might one approach this? I might be missing something obvious as the structure of the question leads me to believe (c) is a fact I can use.
Either I can show the square root of a representation of \(\displaystyle T_{2018}\) isn't rational, or show \(\displaystyle T_{2018}\) is between two consecutive perfect squares?
 
Jun 2013
1,111
590
Lebanon
sequence of unit digits of the $T_i$

\(\displaystyle \{1,2,3,7,3,7,3,7,3,7,3,7,3,7,\text{...}\}\)

so it looks like $T_{2018}$ ends in a $7$
 
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