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x-4-x-3-x-2-x-1-y-2-where-y-is-positive-integer-number-then-find-the-positive-integal-values-of-x-for-which-that-holds-

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Question Number 192793 by York12 last updated on 27/May/23
x^4 +x^3 +x^2 +x+1=y^2 where y is positive integer number then find the positive integal values of (x) for which that holds
$$\boldsymbol{{x}}^{\mathrm{4}} +\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{x}}+\mathrm{1}=\boldsymbol{{y}}^{\mathrm{2}} \:\boldsymbol{{where}}\:\boldsymbol{{y}}\:\boldsymbol{{is}}\:\boldsymbol{{positive}}\:\boldsymbol{{integer}}\:\boldsymbol{{number}} \\ $$$$\boldsymbol{{then}}\:\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{positive}}\:\boldsymbol{{integal}}\:\boldsymbol{{values}}\:\boldsymbol{{of}}\:\left(\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{which}}\:\boldsymbol{{that}}\:\boldsymbol{{holds}} \\ $$
Answered by AST last updated on 27/May/23
4(x^4 +x^3 +x^2 +x+1)=(2y)^2 (2x^2 +x)^2 <(2y)^2 =4y^2 <(2x^2 +x+1)^2 when x>3 ⇒4y^2 cannot be a perfect square when x>3 This implies y^2 cannot also be So,we check x≤3 ⇒(x,y)=(3,11)
$$\mathrm{4}\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\left(\mathrm{2}{y}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{x}^{\mathrm{2}} +{x}\right)^{\mathrm{2}} <\left(\mathrm{2}{y}\right)^{\mathrm{2}} =\mathrm{4}{y}^{\mathrm{2}} <\left(\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} \:{when}\:{x}>\mathrm{3} \\ $$$$\Rightarrow\mathrm{4}{y}^{\mathrm{2}} \:{cannot}\:{be}\:{a}\:{perfect}\:{square}\:{when}\:{x}>\mathrm{3} \\ $$$${This}\:{implies}\:{y}^{\mathrm{2}} \:{cannot}\:{also}\:{be} \\ $$$${So},{we}\:{check}\:{x}\leqslant\mathrm{3} \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\mathrm{3},\mathrm{11}\right) \\ $$
Commented by York12 last updated on 27/May/23
(x^2 +(x/2))^2 <(x^4 +x^3 +x^2 +x+1)<(x^2 +(x/2)+1)^2 for x ∈Even numbers then it is impossible to find a perfect square number Between (x^2 +(x/2))^2 & (x^2 +(x/2)+1)^2 becsuse They are consecuative integers. Then x ∈ odd numbers → (x^2 +(x/2))^2 & (x^2 +(x/2)+1)^2 are not perfect squares but there one perfect square between them : (x^2 +(x/2)+(1/2))^2 and since : (x^2 +(x/2))^2 <(x^4 +x^3 +x^2 +x+1)<(x^2 +(x/2)+1)^2 ∴ x^4 +x^3 +x^2 +x+1=(x^2 +(x/2)+(1/2))^2 x^4 +x^3 +x^2 +x+1=x^4 +x^3 +x^2 +(x^2 /4)+(x/2)+(1/4) x^2 −2x−3=0 → x ∈ {3 ,−1} ∴ x = 3 ( That ′s it )
$$\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} <\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)<\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${for}\:{x}\:\in{Even}\:{numbers}\: \\ $$$${then}\:{it}\:{is}\:{impossible}\:\:{to}\:{find}\:{a}\:\:{perfect}\:{square}\:{number} \\ $$$${Between}\:\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} \:\&\:\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \:{becsuse} \\ $$$${They}\:{are}\:{consecuative}\:{integers}. \\ $$$${Then}\:{x}\:\in\:{odd}\:{numbers}\:\rightarrow\:\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} \:\&\:\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${are}\:{not}\:{perfect}\:{squares}\:{but}\:{there}\:{one}\:{perfect}\:{square} \\ $$$${between}\:{them}\:: \\ $$$$\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${and}\:{since}\:: \\ $$$$\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} <\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)<\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\therefore\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}={x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}=\mathrm{0}\:\rightarrow\:{x}\:\in\:\left\{\mathrm{3}\:,−\mathrm{1}\right\}\: \\ $$$$\therefore\:\:{x}\:=\:\mathrm{3}\:\left(\:{That}\:'{s}\:\:{it}\:\right) \\ $$
Answered by MM42 last updated on 28/May/23
x(x+1)(x^2 +1)=(y−1)(y+1) ∗ the lrft side is always even ,so the right side must also be even. (y−1)(y+1),two even numbers are consecutive we can arrang the left side of the above relationship of the follows (x^2 +x)(x^2 +1) or (x+1)(x^3 +x) or x(x^3 +x^2 +x+1) case 1 ∣x^2 +x−x^2 −1∣=2⇒x=−1 × & x=3 ⇒y=11 ✓ case 2 ∣x^3 +x−x−1∣=2⇒x=−1 & (3)^(1/3) × case 3 ∣x^3 +x^2 +x+1−x∣=2⇒∣x^3 +x^2 +1∣=2 ×(because x>0 ) therefore ,the only answer is ; x=3 & y=11
$${x}\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)=\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right)\:\:\ast \\ $$$${the}\:{lrft}\:{side}\:{is}\:{always}\:{even}\:,{so}\:\:{the}\:{right}\:{side}\: \\ $$$${must}\:{also}\:{be}\:{even}. \\ $$$$\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right),{two}\:{even}\:{numbers}\:{are}\:{consecutive} \\ $$$${we}\:{can}\:{arrang}\:{the}\:{left}\:{side}\:{of}\:{the}\:{above}\:{relationship}\:{of}\:{the}\:{follows} \\ $$$$\left({x}^{\mathrm{2}} +{x}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\:{or}\:\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}\right)\:{or}\:{x}\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$${case}\:\mathrm{1} \\ $$$$\mid{x}^{\mathrm{2}} +{x}−{x}^{\mathrm{2}} −\mathrm{1}\mid=\mathrm{2}\Rightarrow{x}=−\mathrm{1}\:×\:\&\:{x}=\mathrm{3}\:\Rightarrow{y}=\mathrm{11}\:\checkmark \\ $$$${case}\:\mathrm{2} \\ $$$$\mid{x}^{\mathrm{3}} +{x}−{x}−\mathrm{1}\mid=\mathrm{2}\Rightarrow{x}=−\mathrm{1}\:\&\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:×\: \\ $$$${case}\:\mathrm{3} \\ $$$$\mid{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}−{x}\mid=\mathrm{2}\Rightarrow\mid{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{1}\mid=\mathrm{2}\:×\left({because}\:{x}>\mathrm{0}\:\right) \\ $$$${therefore}\:,{the}\:{only}\:{answer}\:{is}\:;\:{x}=\mathrm{3}\:\&\:{y}=\mathrm{11} \\ $$$$ \\ $$
Commented by York12 last updated on 27/May/23
you can subtract 2 also and you will come up with the same results
$${you}\:{can}\:{subtract}\:\mathrm{2}\:{also}\:{and}\:{you}\:{will}\:{come}\:{up} \\ $$$${with}\:{the}\:{same}\:{results} \\ $$$$ \\ $$
Commented by MM42 last updated on 27/May/23
thank you
$${thank}\:{you} \\ $$
Commented by York12 last updated on 28/May/23
you are welcome sir
$${you}\:{are}\:{welcome}\:{sir} \\ $$
Commented by MM42 last updated on 28/May/23
I am a retired teacher .and i am very happy to be member of this community good luck
$${I}\:{am}\:{a}\:{retired}\:{teacher}\:.{and}\:{i}\:{am}\: \\ $$$${very}\:{happy}\:{to}\:{be}\:{member}\:{of}\:{this}\:{community} \\ $$$${good}\:{luck} \\ $$
Commented by York12 last updated on 28/May/23
Nice to meet you sir I am a junior high scholar in grade 10 and I am preparing for The IMO
$${Nice}\:{to}\:{meet}\:{you}\:{sir}\: \\ $$$${I}\:{am}\:{a}\:{junior}\:{high}\:{scholar}\:{in}\:{grade}\:\mathrm{10}\: \\ $$$${and}\:{I}\:{am}\:{preparing}\:{for}\:{The}\:{IMO} \\ $$
Commented by MM42 last updated on 28/May/23
i hpoe you have e good future
$${i}\:{hpoe}\:{you}\:{have}\:{e}\:{good}\:{future} \\ $$
Commented by York12 last updated on 28/May/23
thanks really appreciate it
$${thanks}\:{really}\:{appreciate}\:{it} \\ $$

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