Given a=^3 (√(7+(√(50)))),b=^3 (√(7−(√(50)))).Prove that a^7 +b^7 is an even number.


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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

	$Thank you sir a lot . Please, I get permission$ $to submit a way as follows :$ $i) If n is odd then A = {(p + q \sqrt{r})}^{2 k + 1} + {(p - q \sqrt{r})}^{2 k + 1}$ $= 2 p [{(p + q \sqrt{r})}^{2 k} - {(p + q)}^{2 k - 1} (p - q \sqrt{r}) +$ $. . . {(p - q \sqrt{r})}^{2 k}] \Rightarrow A is even$ $ii) If n = 2 k then A = {[{(p + q \sqrt{r})}^{k} + {(p - q \sqrt{r})}^{k}]}^{2}$ $+ for k = 1 \Rightarrow A = {2 p}^{2} + {2 q}^{2} r \Rightarrow A is even$ $+ Suppose A is even \forall n ⩽ 2 k i . e$ $A_{n} = {(p + q \sqrt{r})}^{n} + {(p - q \sqrt{r})}^{n} is even \forall n ⩽ 2 k$ $Then A_{2 n + 2} = {[{(p + q \sqrt{r})}^{k + 1} + {(p - q \sqrt{r})}^{k + 1}]}^{2}$ $- 2 {(p^{2} - q^{2} r)}^{k + 1} = {(2 M)}^{2} - 2 {(p^{2} - q^{2} r)}^{k + 1} = 2 Q$ $which shows that^{″} A is {even}^{″} true for \forall n = 2 k$ $F rom i) and ii) follows A is even \forall n$ $Sir, is it all right ? Thank you sir!$
	Commented by MJS last updated on 30/Jun/20

	$looks right to me$
	Answered by 1549442205 last updated on 03/Jul/20

	$ab =^{3} \sqrt{(7 + \sqrt{50}) (7 - \sqrt{50})} = - 1 . Hence,$ $a^{3} + b^{3} = 14 = {(a + b)}^{3} - 3 ab (a + b)$ $= {(a + b)}^{3} + 3 (a + b) \Rightarrow {(a + b)}^{3} + 3 (a + b) - 14 = 0$ $\Leftrightarrow (a + b - 2) [{(a + b)}^{2} + 2 (a + b) + 7] = 0$ $\Rightarrow a + b = 2 . From$ $a^{7} + b^{7} = (a + b) (a^{6} - a^{5} b + a^{4} b^{2} - a^{3} b^{3} + a^{2} b^{4} - {ab}^{5} + b^{6})$ $= 2 (a^{6} - a^{5} b + a^{4} b^{2} - a^{3} b^{3} + a^{2} b^{4} - {ab}^{5} + b^{6})$ $which shows that a^{7} + b^{7} is odd number (q . e . d)$ $other way :$ $a^{4} + b^{4} = {(a^{2} + b^{2})}^{2} - 2 a^{2} b^{2} = {[{(a + b)}^{2} - 2 ab]}^{2} - 2 a^{2} b^{2}$ $= {(2^{2} - 2 (- 1))}^{2} - 2 \times 1 = 6^{2} - 2 = 34$ $a^{7} + b^{7} = (a^{3} + b^{3}) (a^{4} + b^{4}) - a^{3} b^{3} (a + b)$ $= 14 \times 34 - {(- 1)}^{3} \times 2 = 478, so a^{7} + b^{7}$ $is a evev number (q . e . d)$
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