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Quaint Quintic Question (Posted on 2012-10-21)

If α = ⁵√(41+29√2) + ⁵√(41-29√2) is real, then show that α must be an integer. No calculators or computer programs allowed.

See The Solution Submitted by K Sengupta

Rating: 3.3333 (3 votes)

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Solution

Comment 1 of 1

Let        a = u + v,   where u = (41 + 29sqrt(2))^1/5, v = (41 – 29sqrt(2))^1/5

then:     uv = (41² – 2*29²)^1/5 = (-1)^1/5 = -1       for real u, v       (1)

a⁵ = (u + v)⁵      = u⁵ + 5u⁴v + 10u³v² + 10u²v³ + 5uv⁴ +v⁵

                        = u⁵ +v⁵ + 5uv(u³ + v³) + 10u²v²(u+v)

                        = u⁵ +v⁵ + 5uv[(u + v)³ – 3uv(u + v)] + 10(uv)²(u+v)

                        = 82 + 5(-1)[a³ – 3(-1)a] + 10(-1)²a     using (1)

                        = 82 – 5a³ - 15a +10a

giving:              a⁵ + 5a³ + 5a – 82 = 0

Factorising:        (a – 2)(a⁴ +2a³ + 9a² + 18a + 41) = 0

We know that there is only one real value for a, so:    a = 2    QED

Posted by Harry on 2012-10-21 22:18:10

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