Find the number of values for m for which (m+i)^4 is an integer? i

Question

#i=sqrt(-1)#

885 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-16T17:00:35

# AA m in {0,+-1}, (m+i)^4 in ZZ.#

Using the Binomial Theorem, we have, #(m+i)^4,#

#=""_4C_0m^4i^0+""_4C_1m^3i+""_4C_2m^2i^2+""_4C_3m^1i^3+""_4C_4m^0i^4,#

#=m^4+4m^3i-6m^2-4mi+1,#

# rArr (m+i)^4=(m^4-6m^2+1) +(4m^3-4m)i.#

In order that #(m+i)^4# be integer (real), equating the Real and

Imaginary parts of both sides, we must have,

#(m^4-6m^2+1) in RR, and 4m^3-4m=4m(m^2-1)=0.#

Now, #4m^3-4m=4m(m^2-1)=0 rArr m=0, m=+-1.#

#"Also, "AA m in {0,+-1}, (m^4-6m^2+1) in RR" holds good."#

# :. m in {0,-1,+1}.#

Hence, there are #3# values of #m# such that, #(m+i)^4 in ZZ.#