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Question Number 204350 by universe last updated on 14/Feb/24

Answered by mr W last updated on 14/Feb/24

$$P\left(a\right)=a^4+a^{2}b+ac+d=1...\left(i\right)$$$$P\left(b\right)={ab}^3+b^3+bc+d=-1...\left(ii\right)$$$$\left(i\right)-\left(ii\right):$$$$a(a^3-b^3)+b(a^2-b^2)+c(a-b)=2$$$$a(a^2+b^2+ab)+b(a+b)+c=\frac{2}{a-b}=wholenumber$$$$\Rightarrow a-b=\pm 1,\pm 2$$$$\Rightarrow \mid a-b{\mid }_{max}=2✓$$

Commented by universe last updated on 14/Feb/24

$$butsirhowitispossiblep\left(b\right)=-1$$$$becausea,b,c,dallarepositiveintegerso$$$${ab}^3+b^3+bc+d⩾0ithinkquestioniswrong??$$

Commented by mr W last updated on 14/Feb/24

$$ithinkthequestionmeansjust$$$$integernumbers\left(\mathbb{Z}\right).$$$$indifferentcountriesitmayhave$$$$differentmeanings.$$

Commented by mr W last updated on 14/Feb/24

Commented by universe last updated on 14/Feb/24

$$thankyousir$$

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