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Question-6593

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Question Number 6593 by Tawakalitu. last updated on 04/Jul/16
Commented by Yozzii last updated on 05/Jul/16
Given that p is prime and p≤100, show that there are solutions (x,y), x,y∈Z, for which y^(37) ≡x^3 +37 (mod p). −−−−−−−−−−−−−−−−−−−−−−− An attempt... (probably misunderstood problem) ⇒∃r∈Z^≥ for p∈[2,100] with y^(37) −rp=x^3 +37 and y,x∈Z. ∴ r=((y^(37) −x^3 −37)/p). r≥0 ∴ y^(37) ≥x^3 +37 y^(37) ≥x^3 +27+10 y^(37) ≥(x+3)(x^2 −3x+9)+10 and there are integers x,y satisfying this inequality. r=((y^(37) −(x+3)(x^2 −3x+9)−10)/p) Suppose p=2. ∴ r=((y^(37) −(x^3 +1))/2)−18. If x is even then x^3 +1 is odd. So, for r∈Z^≥ , y^(37) must be odd so that y^(37) −x^3 −1 is even or ∃k∈Z where y^(37) −x^3 −1=2k⇒r=k−15∈Z. If x is odd then y is even. In both cases even or odd integers x or y can be found accordingly. − − − − − − − − − − All other primes are odd after p=2. r=((y^(37) −(x^3 +1))/p)−18 continue...
$${Given}\:{that}\:{p}\:{is}\:{prime}\:{and}\:{p}\leqslant\mathrm{100}, \\ $$$${show}\:{that}\:{there}\:{are}\:{solutions}\:\left({x},{y}\right),\: \\ $$$${x},{y}\in\mathbb{Z},\:{for}\:{which}\:{y}^{\mathrm{37}} \equiv{x}^{\mathrm{3}} +\mathrm{37}\:\left({mod}\:{p}\right). \\ $$$$−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${An}\:{attempt}…\:\left({probably}\:{misunderstood}\:{problem}\right) \\ $$$$\Rightarrow\exists{r}\in\mathbb{Z}^{\geqslant} \:{for}\:{p}\in\left[\mathrm{2},\mathrm{100}\right]\:{with}\:{y}^{\mathrm{37}} −{rp}={x}^{\mathrm{3}} +\mathrm{37} \\ $$$${and}\:{y},{x}\in\mathbb{Z}. \\ $$$$\therefore\:{r}=\frac{{y}^{\mathrm{37}} −{x}^{\mathrm{3}} −\mathrm{37}}{{p}}.\:{r}\geqslant\mathrm{0}\:\therefore\:{y}^{\mathrm{37}} \geqslant{x}^{\mathrm{3}} +\mathrm{37} \\ $$$${y}^{\mathrm{37}} \geqslant{x}^{\mathrm{3}} +\mathrm{27}+\mathrm{10} \\ $$$${y}^{\mathrm{37}} \geqslant\left({x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{9}\right)+\mathrm{10}\:{and}\:{there} \\ $$$${are}\:{integers}\:{x},{y}\:{satisfying}\:{this}\:{inequality}. \\ $$$${r}=\frac{{y}^{\mathrm{37}} −\left({x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{9}\right)−\mathrm{10}}{{p}} \\ $$$${Suppose}\:{p}=\mathrm{2}.\: \\ $$$$\therefore\:{r}=\frac{{y}^{\mathrm{37}} −\left({x}^{\mathrm{3}} +\mathrm{1}\right)}{\mathrm{2}}−\mathrm{18}. \\ $$$${If}\:{x}\:{is}\:{even}\:{then}\:{x}^{\mathrm{3}} +\mathrm{1}\:{is}\:{odd}. \\ $$$${So},\:{for}\:{r}\in\mathbb{Z}^{\geqslant} ,\:{y}^{\mathrm{37}} \:{must}\:{be}\:{odd}\:{so}\:{that} \\ $$$${y}^{\mathrm{37}} −{x}^{\mathrm{3}} −\mathrm{1}\:{is}\:{even}\:{or}\:\exists{k}\in\mathbb{Z}\:{where}\: \\ $$$${y}^{\mathrm{37}} −{x}^{\mathrm{3}} −\mathrm{1}=\mathrm{2}{k}\Rightarrow{r}={k}−\mathrm{15}\in\mathbb{Z}. \\ $$$${If}\:{x}\:{is}\:{odd}\:{then}\:{y}\:{is}\:{even}.\:{In}\:{both}\:{cases} \\ $$$${even}\:{or}\:{odd}\:{integers}\:{x}\:{or}\:{y}\:{can}\:{be}\:{found}\:{accordingly}. \\ $$$$−\:\:\:\:\:−\:\:\:\:\:−\:\:\:\:−\:\:\:\:−\:\:\:\:−\:\:\:\:−\:\:\:\:−\:\:\:\:−\:\:\:\:−\: \\ $$$${All}\:{other}\:{primes}\:{are}\:{odd}\:{after}\:{p}=\mathrm{2}. \\ $$$${r}=\frac{{y}^{\mathrm{37}} −\left({x}^{\mathrm{3}} +\mathrm{1}\right)}{{p}}−\mathrm{18} \\ $$$${continue}… \\ $$
Commented by prakash jain last updated on 05/Jul/16
y^(37) ≡x^3 +37 (mod p) Since the question is a for limited set of number i am just listing a solution. case p≤37 p=2 37≡1 mod 2 ⇒x=0 y=1 p=3 37≡1 mod 3⇒x=0 y=1 p=5 37≡2 mod 5⇒x=−1 y=1 p=7 37≡2 mod 7⇒x=−1 y=1 p=11 37≡−7 mod 11⇒x=2 y=1 p=13 37≡−2 mod 13⇒x=1 y=1 p=17 37≡3 mod 17⇒Solve y^5 =x^3 +3(mod 17) p=19 37≡−1 mod 19⇒x=1,y=0 p=23 37≡−9 mod 23⇒x=2,y=−1 p=29 37≡8 mod 29⇒x=−2,y=0 p=31 37≡6 mod 31⇒Solve y^7 =x^3 +6(mod 31) p≡37 37≡0 mod 37⇒x=0,y=0
$${y}^{\mathrm{37}} \equiv{x}^{\mathrm{3}} +\mathrm{37}\:\left(\mathrm{mod}\:{p}\right) \\ $$$$\mathrm{Since}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{a}\:\mathrm{for}\:\mathrm{limited}\:\mathrm{set}\:\mathrm{of} \\ $$$$\mathrm{number}\:\mathrm{i}\:\mathrm{am}\:\mathrm{just}\:\mathrm{listing}\:\mathrm{a}\:\mathrm{solution}. \\ $$$$\mathrm{case}\:{p}\leqslant\mathrm{37} \\ $$$$\:\:\:{p}=\mathrm{2}\:\:\:\:\:\mathrm{37}\equiv\mathrm{1}\:\mathrm{mod}\:\mathrm{2}\:\Rightarrow{x}=\mathrm{0}\:{y}=\mathrm{1} \\ $$$$\:\:\:{p}=\mathrm{3}\:\:\:\:\:\mathrm{37}\equiv\mathrm{1}\:\mathrm{mod}\:\mathrm{3}\Rightarrow{x}=\mathrm{0}\:{y}=\mathrm{1} \\ $$$$\:\:\:{p}=\mathrm{5}\:\:\:\:\:\mathrm{37}\equiv\mathrm{2}\:\mathrm{mod}\:\mathrm{5}\Rightarrow{x}=−\mathrm{1}\:{y}=\mathrm{1} \\ $$$$\:\:\:{p}=\mathrm{7}\:\:\:\:\:\mathrm{37}\equiv\mathrm{2}\:\mathrm{mod}\:\mathrm{7}\Rightarrow{x}=−\mathrm{1}\:{y}=\mathrm{1} \\ $$$$\:\:\:{p}=\mathrm{11}\:\:\:\mathrm{37}\equiv−\mathrm{7}\:\mathrm{mod}\:\mathrm{11}\Rightarrow{x}=\mathrm{2}\:{y}=\mathrm{1} \\ $$$$\:\:\:{p}=\mathrm{13}\:\:\:\:\mathrm{37}\equiv−\mathrm{2}\:\mathrm{mod}\:\mathrm{13}\Rightarrow{x}=\mathrm{1}\:{y}=\mathrm{1} \\ $$$$\:\:\:{p}=\mathrm{17}\:\:\:\:\mathrm{37}\equiv\mathrm{3}\:\mathrm{mod}\:\mathrm{17}\Rightarrow\mathrm{Solve}\:{y}^{\mathrm{5}} ={x}^{\mathrm{3}} +\mathrm{3}\left({mod}\:\mathrm{17}\right) \\ $$$$\:\:\:{p}=\mathrm{19}\:\:\:\:\mathrm{37}\equiv−\mathrm{1}\:\mathrm{mod}\:\mathrm{19}\Rightarrow{x}=\mathrm{1},{y}=\mathrm{0} \\ $$$$\:\:\:{p}=\mathrm{23}\:\:\:\mathrm{37}\equiv−\mathrm{9}\:\mathrm{mod}\:\mathrm{23}\Rightarrow{x}=\mathrm{2},{y}=−\mathrm{1} \\ $$$$\:\:\:{p}=\mathrm{29}\:\:\:\mathrm{37}\equiv\mathrm{8}\:\mathrm{mod}\:\mathrm{29}\Rightarrow{x}=−\mathrm{2},{y}=\mathrm{0} \\ $$$$\:\:\:{p}=\mathrm{31}\:\:\mathrm{37}\equiv\mathrm{6}\:\mathrm{mod}\:\mathrm{31}\Rightarrow\mathrm{Solve}\:{y}^{\mathrm{7}} ={x}^{\mathrm{3}} +\mathrm{6}\left({mod}\:\mathrm{31}\right) \\ $$$$\:\:\:{p}\equiv\mathrm{37}\:\:\mathrm{37}\equiv\mathrm{0}\:\mathrm{mod}\:\mathrm{37}\Rightarrow{x}=\mathrm{0},{y}=\mathrm{0} \\ $$

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