If n > 2 and n^4 - 5n^2 + 4 = 20h, is h an integer?

Solution

Step 1: Analyse Question Stem

• n > 2
• \(n^4 – 5n^2 + 4 = 20h\)
• We need to find if h is an integer.

o Now, n will be an integer only if \( n^4 – 5n^2 + 4 = 20h\) is a multiple of 20.
o i.e. h is an integer if \((\frac{(n^4 -5n^2 + 4)}{20})_R= 0\)

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

• According to this statement: n = 4k, where k is an integer.
• So, \((\frac{(n^4 -5n^2 + 4)}{20})_R= (\frac{4^4*k^4 -5*4^2*k^2 + 4}{20})_R = (\frac{256*k^4 -80*k^2 + 4}{20})_R \)
• Depending on the values of k there can be many cases, for example, consider the following two cases:

o Case 1: if k = 5,

 \( (\frac{256*k^4 -80*k^2 + 4}{20})_R =(\frac{256*5^4}{20})_R -(\frac{80*5^2}{20})_R + (\frac{4}{20})_R =0 - 0 + 4 = 4 \), which is not equal to 0.
o Case 2: if k = 1,
o \( (\frac{256*k^4 -80*k^2 + 4}{20})_R =(\frac{256 -80 + 4}{20})_R = 0 \)
• We are getting two contradictory results.

• According to this statement: n = 5m, where m is an integer.
• So, \((\frac{(n^4 -5n^2 + 4)}{20})_R= (\frac{5^4*m^4 -5*5^2*m^2 + 4}{20})_R = (\frac{625*m^4 -125*m^2 + 4}{20})_R \)
• There can be two cases:

o Case 1: if m is even, say m = 2,

 In that case \( (\frac{625*k^4 -125*k^2 + 4}{20})_R =(\frac{625*2^4}{20})_R -(\frac{125*2^2}{20})_R +(\frac{ 4}{20})_R = 0 -0 + 4 = 4 \), which is not equal to 0.
o Case 2: if m is odd, say m = 1,

 \( (\frac{625*m^4 -125*m^2 + 4}{20})_R =(\frac{625 -125 + 4}{20})_R = 4 \), which is not equal to 0.
o Thus, h is not an integer.