～～求解～～

沒錯，obviously, 3a^4 + 8a³b - 12a²b² - 112ab³ - 158b^4 = 0 has solution
　╭
＿│a = 0
　│b = 0
　╰
但這條數學題還未就此完結。第一，此方程式有無限多個解。第二，上述的解不符合「一對‘正’整數解」的要求。

所以要繼續做埋落去！

3a^4 + 8a³b – 12a²b² – 112ab³ – 158b^4 = 0
Suppose a, b ≠ 0,
3((a^4)/(b^4)) + 8(a³/b³) – 12(a²/b²) – 112(a/b) – 158 = 0
3(a/b)^4 + 8(a/b)³ – 12(a/b)² – 112(a/b) – 158 = 0

Let x = a/b, the equation becomes:
3x^4 + 8x³ – 12x² – 112x – 158 = 0

Let f(x) = 3x^4 + 8x³ – 12x² – 112x – 158

Note that 3 has factors ± 1 and ± 3; 158 has factors ± 1, ± 2, ± 79 and ± 158
f(1) = 3 × 1^4 + 8 × 1³ – 12 × 1² – 112 × 1 – 158 = – 271 ≠ 0
∴x – 1 is not a factor.
f(– 1) = 3(– 1)^4 + 8(– 1)³ – 12(– 1)² – 112(– 1) – 158 = – 63 ≠ 0
∴x + 1 is not a factor.
f(2) = 3 × 2^4 + 8 × 2³ – 12 × 2² – 112 × 2 – 158 = – 318 ≠ 0
∴x – 2 is not a factor.
f(– 2) = 3(– 2)^4 + 8(– 2)³ – 12(– 2)² – 112(– 2) – 158 = 2 ≠ 0
∴x + 2 is not a factor.
f(79) = 3 × 79^4 + 8 × 79³ – 12 × 79² – 112 × 79 – 158 = 120710657 ≠ 0
∴x – 79 is not a factor.
f(– 79) = 3(– 79)^4 + 8(– 79)³ – 12(– 79)² – 112(– 79) – 158 = 112839729 ≠ 0
∴x + 79 is not a factor.
f(158) = 3 × 158^4 + 8 × 158³ – 12 × 158² – 112 × 158 – 158 = 1900840962 ≠ 0
∴x – 158 is not a factor.
f(– 158) = 3(– 158)^4 + 8(– 158)³ – 12(– 158)² – 112(– 158) – 158 = 1837767362 ≠ 0
∴x + 158 is not a factor.
f(1/3) = 3(1/3)^4 + 8(1/3)³ – 12(1/3)² – 112(1/3) – 158 = – 589/3 ≠ 0
∴3x – 1 is not a factor.
f(– 1/3) = 3(– 1/3)^4 + 8(– 1/3)³ – 12(– 1/3)² – 112(– 1/3) – 158 = – 3301/27 ≠ 0
∴3x + 1 is not a factor.
f(2/3) = 3(2/3)^4 + 8(2/3)³ – 12(2/3)² – 112(2/3) – 158 = – 6346/27 ≠ 0
∴3x – 2 is not a factor.
f(– 2/3) = 3(– 2/3)^4 + 8(– 2/3)³ – 12(– 2/3)² – 112(– 2/3) – 158 = – 814/9 ≠ 0
∴3x + 2 is not a factor.
f(79/3) = 3(79/3)^4 + 8(79/3)³ – 12(79/3)² – 112(79/3) – 158 = 14195273/9 ≠ 0
∴3x – 79 is not a factor.
f(– 79/3) = 3(– 79/3)^4 + 8(– 79/3)³ – 12(– 79/3)² – 112(– 79/3) – 158 = 34856459/27 ≠ 0
∴3x + 79 is not a factor.
f(158/3) = 3(158/3)^4 + 8(158/3)³ – 12(158/3)² – 112(158/3) – 158 = 653693558/27 ≠ 0
∴3x – 158 is not a factor.
f(– 158/3) = 3(– 158/3)^4 + 8(– 158/3)³ – 12(– 158/3)² – 112(– 158/3) – 158 = 196967698/9 ≠ 0
∴3x + 158 is not a factor.

Hence there are not any rational roots of x, only have irrational roots or/and complex roots of x

i.e. a/b = x_i for i = 1, 2, 3, 4 at which represent the four roots of the equation 3x^4 + 8x³ – 12x² – 112x – 158 = 0
Then a = (x_i)b and b = a/(x_i)

If b is a positive integer, then a should not a positive integer.
If a is a positive integer, then b should not a positive integer.

∴a和b不能同時是正整數

因此，我們不可能找到一對‘正’整數a和b使其滿足方程3a^4 + 8a³b – 12a²b² – 112ab³ – 158b^4 = 0。


注意：雖然我們不可以找到一對‘正’整數a和b使其滿足方程3a^4 + 8a³b – 12a²b² – 112ab³ – 158b^4 = 0，但不代表此方程式完全冇解。實際上，此方程式不但有解，而且有無限多組解，只不過在同一組解不會同時出現兩個正整數而已。

To find the general solution of 3a^4 + 8a³b – 12a²b² – 112ab³ – 158b^4 = 0, we first solve 3x^4 + 8x³ – 12x² – 112x – 158 = 0, i.e. find x_1, x_2, x_3 and x_4. But this process must involve Quartic formula which is shown at http://hk.knowledge.yahoo.com/question/?qid=7007030702434,

After finding x_1, x_2, x_3 and x_4, we will quickly know that a = (x_1)b or a = (x_2)b or a = (x_3)b or a = (x_4)b. At this time, we let b = t, for all t in Ｃ.答案呼之欲出。

The general solution is
　╭
＿│a = (x_1)t
　│b = t
　╰
or
　╭
＿│a = (x_2)t
　│b = t
　╰
or
　╭
＿│a = (x_3)t
　│b = t
　╰
or
　╭
＿│a = (x_4)t
　│b = t
　╰
, for all t in Ｃ

2007-08-05 02:38:14 補充：
感冒唔駛食藥，飲盒仔茶就得啦。　同埋　貓朋　，請你哋唔好用gcd、mod嚟撻人啦！！！唔好將條數搞到咁複雜啦！好似我咁計已經冇問題啦！

唔明點解貓朋的兩行solution 會得唔到最佳解答...
條題目係問正"整"數解, number theory 的題目, 用gcd+mod 係最標準的做法, 而且一．D．都．唔．複．雜, 比paul 的答案要簡單十倍.
貓朋的補充: 你係想說k =/= 0,1 的時候吧?

設 f(a,b)= 3a^4 + 8a³b - 12a²b² - 112ab³ - 158b^4 , 當 f(a,b)=0，求 a,b∈ℤ，使得ab≠0
由於 f(a,b) 是齊次多項式，可假設 gcd(a,b)=1 。
f(a,b)≡3a^4 +0-0-0-0≡a^4 ≡0 (mod 2)
所以 a≡0 (mod 2)
因此，
f(a,b)≡0+0-0-0-158b^4 ≡2b^4 ≡0 (mod 4)
由此得 b≡0 (mod 2) ，但這和 gcd(a,b)=1 矛盾。所以 f(a,b)=0 沒有任整數解使得 ab≠0。


圖片參考：http://i175.photobucket.com/albums/w130/bjoechan2003/My%20Cat%2020070530/DSCN2699.jpg


2007-08-05 23:13:22 補充：
Sorry. 我唔係想撻人。其實我開頭都好似paul2357paul咁計，不過覺得太長所以無答。但剛巧看書學到這招，忍唔住要試試，果然醒神。
其實gcd(a,b)=1表示兩數互素，mod是餘數計算。我其實可以簡單普通的描述，不過習慣咁寫，無辦法啦。
最後，留意
3a^4 + 8a³b - 12a²b² - 112ab³ - 158b^4 =(2^k)q
是沒有任何整數解。(為何？)其中 k 是正整數，q 為奇數。

No. There's no solution.
正整數不含0

(a,b) = (0,0)