What an awkward jump

Using the given equations, we see that

$(a + b)^{2} = 1 \Longrightarrow a^{2} + b^{2} + 2ab = 1 \Longrightarrow 2 + 2ab = 1 \Longrightarrow ab = -\dfrac{1}{2}.$

Next, we have that

$(a^{2} + b^{2})(a + b) = 2 \Longrightarrow a^{3} + b^{3} + ab(a + b) = 2$

$\Longrightarrow a^{3} + b^{3} - \dfrac{1}{2} = 2 \Longrightarrow a^{3} + b^{3} = \dfrac{5}{2},$

and that

$(a^{2} + b^{2})^{2} = 4 \Longrightarrow a^{4} + b^{4} + 2(ab)^{2} = 4$

$\Longrightarrow a^{4} + b^{4} + 2*\dfrac{1}{4} = 4 \Longrightarrow a^{4} + b^{4} = \dfrac{7}{2}.$

We then find that

$(a^{3} + b^{3})(a^{4} + b^{4}) = \dfrac{5}{2} * \dfrac{7}{2} = \dfrac{35}{4}$

$\Longrightarrow a^{7} + b^{7} + a^{3}b^{3}(a + b) = a^{7} + b^{7} + (ab)^{3} = a^{7} + b^{7} - \dfrac{1}{8} = \dfrac{35}{4}$

$\Longrightarrow a^{7} + b^{7} = \dfrac{35}{4} + \dfrac{1}{8} = \dfrac{70}{8} + \dfrac{1}{8} = \boxed{\dfrac{71}{8}}.$

Firstly, $(a+b)^2 = 1 \\ a^2 + 2ab + b^2 = 1 \\ 2 + 2ab = 1 \\ 2ab = -1 \\ ab = -\frac{1}{2}$ Let $S_n = a^n + b^n$ . By multiplying this by $(a+b)$ we get: $\begin{aligned} (a+b)S_n &= (a+b)(a^n + b^n) \\ 1\times S_n &= a^{n+1} + b^{n+1} + ab^n + ba^n \\ S_n &= S_{n+1} + ab(a^{n-1} + b^{n-1}) \\ S_n &= S_{n+1} -\frac{1}{2} S_{n-1} \\ \frac{1}{2}S_{n-1} + S_n &= S_{n+1} \end{aligned}$ From this, we have a simple recurrence formula for $S_n$ . Using the initial $S_1 = 1, S_2 = 2$ , we get

$\begin{aligned} S_3 &= \frac{1}{2} + 2 = \frac{5}{2} \\ S_4 &= \frac{2}{2} + \frac{5}{2} = \frac{7}{2} \\ S_5 &= \frac{19}{4} \\ S_6 &= \frac{26}{4} \\ S_7 &= \frac{71}{8} \end{aligned}$

Therefore $a^7 + b^7 = S_7 = \frac{71}{8}$

Solving the first two equations simulaneously gives a, b = (1+√3)/2, (1-√3)/2. Thus a^7 + b^7 = [(1+√3)/2]^7 + [(1-√3)/2]^7 = 71/8 = 8.875.

a^2+(1-a)^2=2 ,2a^2-2a-1=0,a=(1+or-√3)÷2,b=1-a,so a^7+b7=(1-√3)^7+(1+√3)^7\2^7=8.875=71\8#####