求\( \displaystyle (\frac{x}{2}+\frac{y}{3}+\frac{z}{6})^2=\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{6} \)的所有整數解。
(90高中數學能力競賽 中彰投區試題)
h ttp://www.math.nuk.edu.tw/senpengeu/HighSchool/2002_Taiwan_High_Taichung_02.pdf 連結已失效
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Show that the inequality \( \sqrt{x+1}+\sqrt{2x-3}+\sqrt{50-3x}\le 12 \) holds for all values \( x \in R \) for which the left-hand side is defined.
SOLUTION. By (48) with \( n=3 \), \( u_1=\sqrt{x+1} \),\( u_2=\sqrt{2x-3} \), and \( u_3=\sqrt{50-3x} \)
we have \( (\sqrt{x+1}+\sqrt{2x-3}+\sqrt{50-3x})^2 \le 3(x+1+2x-3+50-3x)=144 \),
from which, upon taking the square root, we obtain the result.
Given an aribitrary prime p, solve the Diophantine equation \( x^4+4^x=p \).(第245頁)
SOLUTION. For any integer \( x<0 \), \( 4^x \) is not an integer, and neither is \( x^4+4^x \).
Thus for any prime p, the given equation has no negative solution.
For \( x=0 \) we have \( 0^4+4^0=1 \), which is not a prime, and for \( x=1 \) we have \( 1^4+4^1=5 \), a prime.
We will now show that for any integer \( x\ge 2 \), the number \( x^4+4^x \) is composite. If \( x=2k \) is even, where \( k \in N \), then
\( x^4+4^x=2^4 k^4+4^{2k}=16(k^4+4^{2(k-1)}) \),
which is a composite number. If \( x=2k+1 \)( \( k \in N \) ) is odd, then
\( x^4+4^x=x^4+4 \cdot 4^{2k}=[x^4+4x^2(2^k)^2+4(2^k)^4]-4x^2(2^k)^2 \)
\( =[x^2+2(2^k)^2]^2-(2x \cdot 2^k)^2 \)
\( =[x^2+2x \cdot 2^k+2(2^k)^2][x^2-2x \cdot 2^k+2(2^k)^2] \)
\( =[(x+2^k)^2+2^{2k}][(x-2^k)^2+2^{2k}] \),
which is again compostie, since \( (x±2^k)^2+2^{2k}\ge 2^{2k} \ge 2^2 >1 \).
To summarize, for \( p=5 \) the given equation has the unique solution \( x=1 \),while there are no solutions for any prime \( p \ne 5 \)
