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Is there some formula to easily guess that:

$7+4\sqrt{3}$ can be factorize to: $(2+ \sqrt{3}) ^2$?

Only looking at $7+4\sqrt{3}$, it's difficult to guess that you can factorize like that.

EDIT: Duplicate of Strategies to denest nested radicals $\sqrt{a+b\sqrt{c}}$ , thanks @Martin_R

ThePhi
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    $(a+b\sqrt{3})^2 = a^2+3b^2+2ab\sqrt{3}$ then compare coefficients – Prometheus Dec 01 '21 at 05:16
  • Check this: https://math.stackexchange.com/q/196155/42969, or this: https://math.stackexchange.com/q/1917494/42969. – Martin R Dec 01 '21 at 05:19
  • You can get a clue from the term $4\sqrt 3$ which is nothing but $2\cdot 2\cdot\sqrt 3$, so we can try to write it in the form $(a\pm b)^2$. – RiverX15 Dec 01 '21 at 05:21
  • @MartinR Thanks, it was exactly what I was looking for! – ThePhi Dec 01 '21 at 05:24
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    $7^2 - 3 \cdot 4^2 = 1,$ so possible factors will be $a+b \sqrt 3$ with $a^2 - 3 b^2 = 1$ as well. The first few pirs with $a,b$ non-negative integers are $(1,0); (2,1) ; (7,4) ; (26,15); (97,56) . $ – Will Jagy Dec 01 '21 at 05:24
  • also $ ( 362 + 209 \sqrt 3) (26-15 \sqrt 3) = 7 + 4 \sqrt 3 $ – Will Jagy Dec 01 '21 at 21:03

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