ข้อสองครับ วิธีทำอาจดูไม่งามนัก หากคิดผิดตรงไหนช่วยบอกนะครับ
2. We assume first that there are integers x,y,z which satisfy following equations system:
(eq1) \(x^6+x^3+x^{3}y+y=(x^{3}+1)(x^{3}+y)=147^{157}=(7^{2}.3)^{157}\)
(eq2) \(x^3+x^{3}y+y^2+y+z^9=(y+1)(x^{3}+y)+z^9=157^{147}\).
Since the product of two odd numbers are odd, we have \(x^{3}+1\) and \(x^{3}+y\) as odd numbers,
that is \(x^3\) is even and, from our assumption, x is even. Also y and z are odd.
We use then the fact that \(x^3\equiv0\pmod4\), \(y\equiv\pm1\pmod4\) and \(z\equiv\pm1\pmod4\) to verify each case as follows
(following calculations are in mod 4):
equation 2:
$z\equiv1\pmod4$, $y\equiv1\pmod4$ yields $2\cdot1+1\equiv1$ (wrong)
$z\equiv1\pmod4$, $y\equiv-1\pmod4$ yields $(-1)\cdot0+1\equiv1$ (right)
$z\equiv-1\pmod4$, $y\equiv1\pmod4$ yields $2\cdot1-1\equiv1$ (right)
$z\equiv-1\pmod4$, $y\equiv-1\pmod4$ yields $2\cdot0-1\equiv1$ (wrong),
then set the right 'answer sets' in equation 1:
\(1\cdot(\pm1)\equiv(-1)^{157}\equiv-1\),
which give us \(y\equiv-1\pmod4\).
From equation 1 we can also assume that \(x^3+1={7^{\alpha_{1}}.3^{\beta_{1}}}\) and \(x^3+y={7^{\alpha_{2}}.3^{\beta_{2}}}\), where all exponents are integers.
We calculate each factor again in mod 4, which yields:
\(x^3+1\equiv(-1)\cdot{\alpha_{1}}+(-1).{\beta_{1}}=-{\alpha_{1}}-{\beta_{1}}\equiv1\),
and \(x^3+y\equiv\ (-1)\cdot{\alpha_{2}}+(-1).{\beta_{2}}=-{\alpha_{2}}-{\beta_{2}}\equiv-1\).
That means \({\alpha_{1}}+{\beta_{1}}+{\alpha_{2}}+{\beta_{2}}\equiv0\ne-1=157\cdot3\pmod4\),
which yield the desired contradiction, hence complete the proof. (๕๕๕ ซะเมื่อไหร่
)
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Phew... คงมีวิธีที่ง่ายกว่านี้นะ เพราะอาจคำนวณเศษผิดได้ง่ายมากๆ
Edit: มาแก้ TeX syntax ครับ กระทู้นี้ตอนผมเข้ามาที่นี่ใหม่ๆเลยนะนั่น