Se quiere determinar el valor de los números enteros $a$ y $b$ tales que $$(1+\sqrt{2})^{12}=a+b\,\sqrt{2}$$
Denotemos $x=1+\sqrt{2}$. Entonces,
$x-1=\sqrt{2}$
$(x-1)^2=(\sqrt{2})^2$
$(x-1)^2=2$
$x^2-2x+1=2$
$x^2=2x+1 \quad (1)$
La cantidad pedida $(1+\sqrt{2})^{12}$ es, por tanto,
$x^{12}$
$=(x^2)^6$, y por $(1)$ podemos escribirlo como
$=(2x+1)^6$
$=(2x+1)^6$
$=(4x^2+4x+1)^6$
$\overset{(1)}{=}(4\,(2x+1)+4x+1)^6$
$\overset{(1)}{=}(8x+4+4x+1)^6$
$=(12x+5)^6$
$=\left((12x+5)^2\right)^3$
$=\left(144x^2 +120x +25\right)^3$
$\overset{(1)}{=}\left(144\,(2x+1) +120x +25\right)^3$
$=\left(288\,x+144 +120x +25\right)^3$
$=\left( 408\,x + 169\right)^3$
$=\left( 408\,x + 169\right)^2\cdot \left( 408\,x + 169\right)$
$=\left( 408^2\,x^2 + 2\cdot 408\cdot 169\,x + 169^2\right)\cdot \left( 408\,x + 169\right)$
$=\left( 166\,464\,x^2 + 137\,904\,x + 28\,561\right)\cdot \left( 408\,x + 169\right)$
$\overset{(1)}{=}\left( 166\,464\cdot (2x+1) + 137\,904\,x + 28\,561\right)\cdot \left( 408\,x + 169\right)$
$\overset{(1)}{=}\left( 166\,464\cdot 2x+166\,464 + 137\,904\,x + 28\,561\right)\cdot \left( 408\,x + 169\right)$
$=\left( 332\,928\,x+166\,464 + 137\,904\,x + 28\,561\right)\cdot \left( 408\,x + 169\right)$
$=\left( 470\,832\,x+195\,025 \right)\cdot \left( 408\,x + 169\right)$
$=470\,832\cdot 408\,x^2+(195\,025\cdot 408+470\,832\cdot 169)\,x+195\,025\cdot 169$
$=192\,099\,456\,x^2+159\,140\,808\,x+32\,959\,225$
$\overset{(1)}{=}192\,099\,456\cdot (2x+1)+159\,140\,808\,x+32\,959\,225$
$=192\,099\,456\cdot 2x+ 192\,099\,456 +159\,140\,808\,x+32\,959\,225$
$=384\,198\,912\,x+ 192\,099\,456 +159\,140\,808\,x+32\,959\,225$
$=543\,339\,720\,x+ 225\,058\,681$
$=543\,339\,720\cdot (1+\sqrt{2})+ 225\,058\,681$
$=543\,339\,720+543\,339\,720\,\sqrt{2}+ 225\,058\,681$
$=768\,398\,401+543\,339\,720\,\sqrt{2} \therefore a=768\,398\,401;\,b=543\,339\,720$
$\diamond$
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