0

From this question, one can see that the sum of two irrationals can yield an integer. Specifically a Lucas number can be expressed as

$$ L_n = \alpha^n + \beta^n, $$

for $\alpha=(1+\sqrt{5})/2$ and $\beta=(1-\sqrt{5})/2$ with $n$ integer (the Fibonacci numbers are somehow complementary to this, so I will not count it as an answer). I was just wondering, are there other "famous" numbers such that their sum is an integer, i.e. that the sum of two irrationals yields an even number?

jjagmath
  • 18,214
user2820579
  • 2,389
  • There's a lot of them. You can consider, for example $(1+\sqrt{2})^n + (1-\sqrt{2})^n$. Do you see how you can generalize this? – jjagmath Apr 16 '21 at 16:40
  • This is too vague. Any time you have a quadratic of the form $x^2+2nx+m$ which does not factor over $\mathbb Q$, the roots pass your test. As do $\pi$ and $-\pi$, for example. Or $e, 2n-e$. – lulu Apr 16 '21 at 16:41
  • Perhaps I should add that one of the numbers does not go to zero for $n$ big. – user2820579 Apr 16 '21 at 16:42
  • Easy to find quadratic examples that meet that condition as well. Just pick your favorite quadratic irrational with norm $>1$ and consider $\alpha^n, \overline {\alpha}^n$. – lulu Apr 16 '21 at 16:45
  • I don't understand what you say... – user2820579 Apr 16 '21 at 16:46
  • Say your favorite quadratic irrational is $3+2i$. Then $a_n=(3+2i)^n+(3-2i)^n$ is always an even integer. – lulu Apr 16 '21 at 16:49
  • Mmm but these are not real... – user2820579 Apr 16 '21 at 16:50
  • 1
    You didn't specify that. If your favorite quadratic irrational is $3+2\sqrt {7}$ then $a_n=(3+2\sqrt {7})^n+(3-2\sqrt 7)^n$ is always an even integer. – lulu Apr 16 '21 at 16:52
  • 1
    See here and here for simple proofs that make obvious the quoted result (and its natural generalization). – Bill Dubuque Apr 16 '21 at 17:10

1 Answers1

1

It is very general let $p,q$ be two roots of $$Ax^2+Bx+c=0 \implies Ap^2+Bp+C $$ $$\implies A C_1p^{n+2}+ B C_1 p^{n+1}+C C_1p^n=0~~~(1)$$ Similarly, we have $$AC_1q^{n+2}+ B C_2 q^{n+1}+ C C_2 q^n=0~~~~(2)$$ Adding (1) and (2), we get $$A f_{n+2}+Bf_{n+1}+Cf_{n}=0,$$ where $$f_n=C_1 p^n+ C_2 q^{n}.$$

Z Ahmed
  • 43,235