Question

If n is a prime number greater than 5, show that \[{{n}^{4}}-1\] is divisible by 240.

Answer 1

Hint: At first factorize 240 as in factors 16, 3, 5 and then use the modulus method to show that \[{{n}^{4}}-1\] is divisible by 16, 3, 5 separately henceforth divisible by 240.

Complete step-by-step answer:
In the question we are given that if n is a prime number greater than 5 then we have to prove that \[{{n}^{4}}-1\] is divisible by 240.
We are given that n is an odd number as every prime number greater than 5 is an odd number.
So we can represent n as 2k + 1 where k is any integer greater than 2.
Now we are given an expression \[{{n}^{4}}-1\], now instead of n we will put, n = 2k + 1.
We have to prove that \[{{n}^{4}}-1\] is divisible by 240.
Now at first we will factorize 240 as,
\[\begin{align}
  & 2\left| \!{\underline {\,
  240 \,}} \right. \\
 & 2\left| \!{\underline {\,
  120 \,}} \right. \\
 & 2\left| \!{\underline {\,
  60 \,}} \right. \\
 & 2\left| \!{\underline {\,
  30 \,}} \right. \\
 & 3\left| \!{\underline {\,
  15 \,}} \right. \\
 & \left| \!{\underline {\,
  5 \,}} \right. \\
\end{align}\]
Now we can write 240 as,
\[{{2}^{4}}\times 3\times 5\] or \[16\times 3\times 5\].
If we prove that \[{{n}^{4}}-1\] is divisible by 16, 3, 5 separately then we can say that it is divisible 240.
This can be only if factors are co – prime to each or share no prime factor other than 1 and here 16, 3, 5 are co – prime to each other.
Now as we took n = 2k + 1 so we can write \[{{n}^{4}}-1\] as \[{{\left( 2k+1 \right)}^{4}}-1\]. Hence we will expand using formula \[{{\left( x+1 \right)}^{4}}\] which is equal to \[{{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x+1\] and we will take x as 2k so we get,
\[{{\left( 2k+1 \right)}^{4}}-1={{\left( 2k \right)}^{4}}+4{{\left( 2k \right)}^{3}}+6{{\left( 2k \right)}^{2}}+4\left( 2k \right)+1-1\]
So, on solving we get,
\[16{{k}^{4}}+32{{k}^{3}}+24{{k}^{2}}+8k\]
The whole expression is divisible by 16 if and only if the last two terms \[24{{k}^{2}}+8k\] is divisible by 16.
We can write \[24{{k}^{2}}+8k\] as 8k (3k + 1).
If k is even then 8k (3k + 1) will be divisible by 16 as k will contribute a factor of 2.
If k is odd then 8k (3k + 1) will be divisible by 16 as 3k + 1 will contribute a factor of 2.
So, we can say \[{{n}^{4}}-1\] is divisible by 16.
The expression of \[{{n}^{4}}-1\] in terms of k was \[16{{k}^{4}}+32{{k}^{3}}+24{{k}^{2}}+8k\].
If k is divided by 3 then the possible remainders are 0, 1, 2. So, we can represent it as,
k = 0 (mod 3)
k = 1 (mod 3)
k = 2 (mod 3)
Now for \[{{1}^{st}}\] case k = 0 (mod 3) the expression \[16{{k}^{4}}+32{{k}^{3}}+24{{k}^{2}}+8k\] will be divisible by 3.
Now for \[{{2}^{nd}}\] case k = 1 (mod 3), if k = 1 (mod 3) then (2k + 1) = 0 (mod 3) which means that n is divisible by 3 as n = 2k + 1 which is not possible hence this case can’t be taken.
Now for \[{{3}^{rd}}\] case k = 1 (mod 3) if k = 1 (mod 3) then (2k + 1) = 2 (mod 3) or n = 2 (mod 3).
If n = 2 (mod 3) then, \[{{n}^{4}}\] = 16 (mod 3) or \[{{n}^{4}}\] = 1 (mod 3), (\[{{n}^{4}}\] - 1) = 0 (mod 3)
Hence \[{{n}^{4}}\] - 1 is divisible by 3 by taking all the cases.
So now we will check divisibility of 5 for \[{{n}^{4}}\] - 1.
If n is divisible by 5 it will give remainders 1, 2, 3, 4 as n is a prime and not divisible by 5.
So, we can write n as,
n = 1 (mod 5)
n = 2 (mod 5)
n = 3 (mod 5)
n = 4 (mod 5)
Now let’s consider first case,
n = 1 (mod 5) which can be written as \[{{n}^{4}}={{1}^{4}}\] (mod 5) or \[{{n}^{4}}=1\] (mod 5) or (\[{{n}^{4}}\] - 1) = 0 (mod 5). Hence, it is divisible by 5.
Now let’s consider third case,
n = 3 (mod 5) which can be written as \[{{n}^{4}}={{3}^{4}}\] (mod 5) or \[{{n}^{4}}\] = 81 (mod 5) or \[{{n}^{4}}=1\] (mod 5) or \[{{n}^{4}}\] - 1 = 0 (mod 5).
Hence it is divisible by 5.
Now let’s consider \[{{4}^{th}}\] case,
n = 4 (mod 5) which can be written as \[{{n}^{4}}={{4}^{4}}\] (mod 5) or \[{{n}^{4}}\] = 256 (mod 5) or \[{{n}^{4}}\] = 1 (mod 5) or \[{{n}^{4}}\] - 1 = 0 (mod 5).
Hence it is divisible by 5.
So, combining all the cases we can say that \[{{n}^{4}}\] - 1 is divisible by 5.
Hence, (\[{{n}^{4}}\] - 1) is divisible by 16, 3 and 5.
Hence, proved that \[{{n}^{4}}\] - 1 is divisible by 240.

Note: If a number ‘a’ is divided by ‘b’ and leaves remainder ‘c’ so we represent it by a = c (mod b).
There are some properties of it like if k is any integer we can write a = c (mod b) as
k + a = k + c (mod b).
ak = ck (mod b).
a – k = c – k (mod b).