Find closest integer to $(3+\sqrt7)^4$ by hand

Question

Find the closest integer to $$(3+\sqrt7)^4$$ by hand, without knowing the correct value of $\sqrt7$ (Maybe just knowing that $2<\sqrt7<3$).

My work:$$(3+\sqrt7)^4 = (16+6\sqrt7)^2 = 508+192\sqrt7$$

The "influence" of the uncertain $\sqrt7$ is pretty big if we just expand it out. Please help!

See https://math.stackexchange.com/q/2254817/42969 for a very similar question. — Martin R, Mar 01 '22 at 12:11
What does it mean to "know the correct value of $\sqrt{7}$"? — Carla only proves trivial prop, Mar 01 '22 at 12:11
Helps to note that the minimal polynomial of $3+\sqrt 7$ is $x^2-6x+2$, and that the other root is small. — lulu, Mar 01 '22 at 12:15

score 4 · Accepted Answer · answered Mar 01 '22 at 12:26

4

If you know that $2< \sqrt{7}<3$, then you know that $3-\sqrt{7}$ is less than one. So $(3-\sqrt{7})^4$ is even smaller. So if you add

$$(3+\sqrt{7})^4 + (3-\sqrt{7})^4$$

you don't change the number by very much, but there are no $\sqrt{7}$'s in the result.

answered Mar 01 '22 at 12:26

B. Goddard

Related: $A_n = (3+\sqrt{7})^n + (3-\sqrt{7})^n$ satisfies a second-order linear recurrence with integer coefficients. Find that recurrence and use it to comute $A_4$. You may think of this if you know the corresponding thing for Fibonacci numbers. – GEdgar Mar 01 '22 at 12:35
Is this essentially different from (e.g.) https://math.stackexchange.com/a/2254856/42969, which was already pointed out as a possible duplicate target? – Martin R Mar 01 '22 at 13:06

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