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[alpercay]

$\sqrt{\dfrac{3^8+5^8+34^4}2} $ sayısını hesaplayınız (Çok ünlü bir üniversitenin giriş sınavında sorulduğu belirtilmiş)

Matkafasından Doğan Dönmez hocanın sorduğu bir soru.

[geo]

$\dfrac{9^4+25^4+34^4}{2}=\dfrac{9^4+25^4+(9+25)^4}{2}$

$x=9$, $y=25$ olsun.

$\begin{array}{lcl}
\dfrac{x^4+y^4+(x+y)^4}{2}&=&\dfrac{x^4+y^4+(x^4+4x^3y+6x^2y^2+4xy^3+y^4)}{2}\\
&=& x^4+y^4+2x^3y+3x^2y^2+2xy^3 \\
&=& (x^2+y^2)^2+2xy(x^2+y^2)+x^2y^2 \\
&=& (x^2+y^2+xy)^2 \\
&=& ((x+y)^2-xy)^2 \\
&=& ((9+25)^2-9\cdot 25)^2 \\
&=& (34^2-15^2)^2 \\
&=& (19\cdot 49)^2 \\
&=& 931^2

\end{array}$

$\sqrt{\dfrac{3^8+5^8+34^4}2}=931$

[geo]

$x=3$, $y=5$ ise $34=x^2+y^2$.

$\begin{array}{lcl}
\dfrac{3^8+5^8+34^4}2 &=& \dfrac{x^8+y^8+(x^2+y^2)^4}{2}\\
&=& \dfrac{x^8+y^8+x^8+4x^6y^2+6x^4y^4+4x^2y^6+y^8}2\\
&=& x^8+y^8+2x^6y^2 + 3x^4y^4+2x^2y^6\\
&=& (x^4+y^4)^2 +2x^2y^2(x^4+y^4)+x^4y^4\\
&=& (x^4+y^4+x^2y^2)^2\\
&=& ((x^2+y^2)^2-x^2y^2)^2\\
\sqrt{\dfrac{3^8+5^8+34^4}2}&=& (x^2+y^2-xy)(x^2+y^2+xy)\\
&=& (9+25-15)(9+25+15)\\
&=& (20-1)(50-1)\\
&=& 20\cdot 50 - 20 - 50 - 1 \\
&=& 1000-70+1\\
&=& 931
\end{array}$

[alpercay]

\begin{align*} 3^8+5^8+34^4&=9^4+25^4+(9+25)^4\\ &=9^4+(16+9)^4+(9+16+9)^4\\ &=9^4+(2\cdot 9+7)^4+(3\cdot 9+7)^4\\ &=9^4\cdot (1+(2+a)^4+(1+2+a)^4)\ ,\ a=7/9\\ &=9^4\cdot (1+a^4+(1+a)^4)\, , \,b=2+a=25/9\\ &=9^4\cdot \left(1+b^4+b^4+4b^3+6b^2+4b+1\right)\\ &=9^4\cdot \left(2+2b^4+4b^3+6b^2+4b\right)\\ \dfrac{3^8+5^8+34^4}{2}&=9^4\cdot \left(1+b^4+2b^3+3b^2+2b\right)\\ &\ldots \end{align*}

$1+b^4+2b^3+3b^2+2b=b^4+b^2+1+2(b^3+b^2+b)=(b^2+b+1)^2$  $$\sqrt{\dfrac{3^8+5^8+34^4}{2}}=81(625/81+25/9+81/81)=625+225+81=931$$