Given-a-3-7-50-b-3-7-50-Prove-that-a-7-b-7-is-an-even-number-

Question Number 101052 by 1549442205 last updated on 30/Jun/20

Given a =^{3} \sqrt{7 + \sqrt{50}}, b =^{3} \sqrt{7 - \sqrt{50}} . Prove

that a^{7} + b^{7} is an even number .

Answered by MJS last updated on 30/Jun/20

a = \sqrt[3]{7 + 5 \sqrt{2}} = 1 + \sqrt{2}

b = \sqrt[3]{7 - 5 \sqrt{2}} = 1 - \sqrt{2}

we can show that {(p + q \sqrt{r})}^{n} + {(p - q \sqrt{r})}^{n}

with p, q \in Z and r, n \in N must always be

an even number

α = p; β = q \sqrt{r}

{(α + β)}^{n} + {(α - β)}^{n} leads to cancelling out

some terms and doubling the remaining

terms \Rightarrow {(α + β)}^{n} + {(α - β)}^{n} = 2 k

I {haven}^{'} t got the time to type the proof but

{it}^{'} s obvious

Commented by 1549442205 last updated on 01/Jul/20

Thank you sir a lot . You are welcome .

Commented by 1549442205 last updated on 30/Jun/20

Thank you sir a lot . Please, I get permission

to submit a way as follows :

i) If n is odd then A = {(p + q \sqrt{r})}^{2 k + 1} + {(p - q \sqrt{r})}^{2 k + 1}

= 2 p [{(p + q \sqrt{r})}^{2 k} - {(p + q)}^{2 k - 1} (p - q \sqrt{r}) +

Missing or unrecognized delimiter for \left

ii) If n = 2 k then A = {[{(p + q \sqrt{r})}^{k} + {(p - q \sqrt{r})}^{k}]}^{2}

+ for k = 1 \Rightarrow A = {2 p}^{2} + {2 q}^{2} r \Rightarrow A is even

+ Suppose A is even \forall n ⩽ 2 k i . e

A_{n} = {(p + q \sqrt{r})}^{n} + {(p - q \sqrt{r})}^{n} is even \forall n ⩽ 2 k

Then A_{2 n + 2} = {[{(p + q \sqrt{r})}^{k + 1} + {(p - q \sqrt{r})}^{k + 1}]}^{2}

- 2 {(p^{2} - q^{2} r)}^{k + 1} = {(2 M)}^{2} - 2 {(p^{2} - q^{2} r)}^{k + 1} = 2 Q

which shows that^{″} A is {even}^{″} true for \forall n = 2 k

F rom i) and ii) follows A is even \forall n

Sir, is it all right ? Thank you sir!

Commented by MJS last updated on 30/Jun/20

looks right to me

Answered by 1549442205 last updated on 03/Jul/20

ab =^{3} \sqrt{(7 + \sqrt{50}) (7 - \sqrt{50})} = - 1 . Hence,

a^{3} + b^{3} = 14 = {(a + b)}^{3} - 3 ab (a + b)

= {(a + b)}^{3} + 3 (a + b) \Rightarrow {(a + b)}^{3} + 3 (a + b) - 14 = 0

\Leftrightarrow (a + b - 2) [{(a + b)}^{2} + 2 (a + b) + 7] = 0

\Rightarrow a + b = 2 . From

a^{7} + b^{7} = (a + b) (a^{6} - a^{5} b + a^{4} b^{2} - a^{3} b^{3} + a^{2} b^{4} - {ab}^{5} + b^{6})

= 2 (a^{6} - a^{5} b + a^{4} b^{2} - a^{3} b^{3} + a^{2} b^{4} - {ab}^{5} + b^{6})

which shows that a^{7} + b^{7} is odd number (q . e . d)

other way :

a^{4} + b^{4} = {(a^{2} + b^{2})}^{2} - 2 a^{2} b^{2} = {[{(a + b)}^{2} - 2 ab]}^{2} - 2 a^{2} b^{2}

= {(2^{2} - 2 (- 1))}^{2} - 2 \times 1 = 6^{2} - 2 = 34

a^{7} + b^{7} = (a^{3} + b^{3}) (a^{4} + b^{4}) - a^{3} b^{3} (a + b)

= 14 \times 34 - {(- 1)}^{3} \times 2 = 478, so a^{7} + b^{7}

is a evev number (q . e . d)