Answer to Question #102337 in Algebra for Glory okoli

Question

Let a and b be the roots of x²-14x+36=0 show that a^n+b^n is divisible by 2n

Answer 1

Solution. Solve the quadratic equation

$$x^2-14x+36=0$$

$$D=(-14)^2-4\times36=52$$

The roots of the equation is equal to

$$x_1=\frac{14+\sqrt{52}}{2}=7+\sqrt{13}$$

$$x_2=\frac{14-\sqrt{52}}{2}=7-\sqrt{13}$$

Consider the value

$$(7+\sqrt{13})^k+(7-\sqrt{13})^k=$$

$$=(7+\sqrt{13}+7-\sqrt{13})((7+\sqrt{13})^{k-1}+(7-\sqrt{13})^{k-1})-$$

$$-(7+\sqrt{13})(7-\sqrt{13})^{k-1}-(7-\sqrt{13})(7+\sqrt{13})^{k-1}=$$

$$=14((7+\sqrt{13})^{k-1}+(7-\sqrt{13})^{k-1})-36((7+\sqrt{13})^{k-2}+(7-\sqrt{13})^{k-2})$$

For the sum we got the recurrence equation

$$S_k=14S_{k-1}-36S_{k-2}$$

Solve the recurrence equation.

$$S_1=7+\sqrt{13}+7-\sqrt{13}=14$$

$$S_2=2\times49+14\sqrt{13}-14\sqrt{13}+2\times13=124$$

S1 is divisible by 2; S2 is divisible by 4.

Find S3

$$S_3=14\times S_2-36\times S_1=14\times124-36\times14=1232$$

1232 is not divisible by 6.