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Question Number 4423 by Rasheed Soomro last updated on 24/Jan/16

Determine f(t) such that ∀ t∈Z x^2 =8x+f(t) has integer-solution.

$$\mathrm{Determine}\:\mathrm{f}\left(\mathrm{t}\right)\:\mathrm{such}\:\mathrm{that}\:\forall\:\mathrm{t}\in\mathbb{Z} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{8x}+\mathrm{f}\left(\mathrm{t}\right)\:\mathrm{has}\:\mathrm{integer}-\mathrm{solution}. \\ $$

Commented by Yozzii last updated on 24/Jan/16

Try f(t)=c; c is a real constant. x^2 −8x−c=0 x=((8±(√(64+4c)))/2) x=((8±2(√(16+c)))/2) x=4±(√(16+c)) For x∈Z⊂R⇒16+c≥0 ∧ 16+c=n^2 for ∀ n∈Z^+ +{0}. Hence f(t)=c for integer solutions x to exist for x^2 =8x+f(t) if c=n^2 −16 for ∀n∈Z. Checking: x^2 =8x+n^2 −16 x^2 −8x+16−n^2 =0 ⇒ x=((8±(√(64−64+4n^2 )))/2) x=((8±2n)/2) x=4±n∈Z.

$${Try}\:{f}\left({t}\right)={c};\:{c}\:{is}\:{a}\:{real}\:{constant}. \\ $$$${x}^{\mathrm{2}} −\mathrm{8}{x}−{c}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{8}\pm\sqrt{\mathrm{64}+\mathrm{4}{c}}}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{8}\pm\mathrm{2}\sqrt{\mathrm{16}+{c}}}{\mathrm{2}} \\ $$$${x}=\mathrm{4}\pm\sqrt{\mathrm{16}+{c}} \\ $$$${For}\:{x}\in\mathbb{Z}\subset\mathbb{R}\Rightarrow\mathrm{16}+{c}\geqslant\mathrm{0}\:\wedge\:\mathrm{16}+{c}={n}^{\mathrm{2}} \\ $$$${for}\:\forall\:{n}\in\mathbb{Z}^{+} +\left\{\mathrm{0}\right\}.\:{Hence}\:{f}\left({t}\right)={c} \\ $$$${for}\:{integer}\:{solutions}\:{x}\:{to}\:{exist}\:{for} \\ $$$${x}^{\mathrm{2}} =\mathrm{8}{x}+{f}\left({t}\right)\:{if}\:{c}={n}^{\mathrm{2}} −\mathrm{16}\:{for}\:\forall{n}\in\mathbb{Z}. \\ $$$$ \\ $$$${Checking}:\:{x}^{\mathrm{2}} =\mathrm{8}{x}+{n}^{\mathrm{2}} −\mathrm{16} \\ $$$${x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{16}−{n}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{x}=\frac{\mathrm{8}\pm\sqrt{\mathrm{64}−\mathrm{64}+\mathrm{4}{n}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{8}\pm\mathrm{2}{n}}{\mathrm{2}} \\ $$$${x}=\mathrm{4}\pm{n}\in\mathbb{Z}. \\ $$$$ \\ $$$$ \\ $$

Commented by Yozzii last updated on 24/Jan/16

x^2 −8x−f(t)=0 x=((8±(√(64+4f(t))))/2)=4±(√(16+f(t))). f(t) , ∀t∈Z,must satisfy (1) f(t)≥−16 and (2) f(t)+16=N^2 where N∈Z. Thus, let t=N⇒f(t)=t^2 −16. This further suggests that f(t)=(P(t))^2 −16 where P(t) is a polynomial in t with real integer coefficients generally. ∴ x=4±(√((P(t))^2 −16+16))=4±P(t). So x=4±P(t)∈Z if t∈Z.

$${x}^{\mathrm{2}} −\mathrm{8}{x}−{f}\left({t}\right)=\mathrm{0} \\ $$$${x}=\frac{\mathrm{8}\pm\sqrt{\mathrm{64}+\mathrm{4}{f}\left({t}\right)}}{\mathrm{2}}=\mathrm{4}\pm\sqrt{\mathrm{16}+{f}\left({t}\right)}. \\ $$$${f}\left({t}\right)\:,\:\forall{t}\in\mathbb{Z},{must}\:{satisfy}\:\left(\mathrm{1}\right)\:{f}\left({t}\right)\geqslant−\mathrm{16} \\ $$$${and}\:\left(\mathrm{2}\right)\:{f}\left({t}\right)+\mathrm{16}={N}^{\mathrm{2}} \:\:{where}\:{N}\in\mathbb{Z}. \\ $$$${Thus},\:{let}\:{t}={N}\Rightarrow{f}\left({t}\right)={t}^{\mathrm{2}} −\mathrm{16}. \\ $$$${This}\:{further}\:{suggests}\:{that}\:{f}\left({t}\right)=\left({P}\left({t}\right)\right)^{\mathrm{2}} −\mathrm{16} \\ $$$${where}\:{P}\left({t}\right)\:{is}\:{a}\:{polynomial}\:{in}\:{t}\:{with}\:{real} \\ $$$${integer}\:{coefficients}\:{generally}. \\ $$$$\therefore\:{x}=\mathrm{4}\pm\sqrt{\left({P}\left({t}\right)\right)^{\mathrm{2}} −\mathrm{16}+\mathrm{16}}=\mathrm{4}\pm{P}\left({t}\right). \\ $$$${So}\:{x}=\mathrm{4}\pm{P}\left({t}\right)\in\mathbb{Z}\:{if}\:{t}\in\mathbb{Z}. \\ $$

Commented by Rasheed Soomro last updated on 24/Jan/16

G^(O^( v) O) D!

$$\mathrm{G}^{\mathcal{O}^{\:\mathrm{v}} \mathcal{O}} \mathrm{D}! \\ $$

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