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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