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Be-p-a-prime-number-arbitrary-Solve-on-positive-integers-x-y-z-xy-z-2-3p-4-x-yz-2-p-4-

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Question Number 160711 by HongKing last updated on 05/Dec/21
Be p a prime number , arbitrary. Solve on positive integers (x;y;z) { ((xy + z^2 = 3p + 4)),((x + yz^2 = p + 4)) :}
$$\mathrm{Be}\:\:\boldsymbol{\mathrm{p}}\:\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:,\:\mathrm{arbitrary}. \\ $$$$\mathrm{Solve}\:\mathrm{on}\:\mathrm{positive}\:\mathrm{integers}\:\:\left(\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}};\boldsymbol{\mathrm{z}}\right) \\ $$$$\begin{cases}{\mathrm{xy}\:+\:\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{3p}\:+\:\mathrm{4}}\\{\mathrm{x}\:+\:\mathrm{yz}^{\mathrm{2}} \:=\:\mathrm{p}\:+\:\mathrm{4}}\end{cases} \\ $$
Answered by mr W last updated on 06/Dec/21
(i)−(ii): x(y−1)+z^2 (1−y)=2p (x−z^2 )(y−1)=2p= { ((1×2p)),((2×p)) :} y−1=1, x−z^2 =2p y=2, x=z^2 +2p z^2 +2p+2z^2 =p+4 3z^2 =4−p ⇒not possible, since LHS=odd ≥3, RHS=even≤2 y−1=2p, x−z^2 =1 y=2p+1, x=z^2 +1 z^2 +1+(2p+1)z^2 =p+4 2(p+1)z^2 =p+3 (p+1)(2z^2 −1)=2 ⇒not possible, since LHS≥3 y−1=2, x−z^2 =p y=3, x=z^2 +p z^2 +p+3z^2 =p+4 z^2 =1 ⇒z=1, x=1+p ✓ y−1=p, x−z^2 =2 y=p+1, x=z^2 +2 z^2 +2+(p+1)z^2 =p+4 z^2 =1 ⇒z=1, x=3, y=p+1 ✓ summary of solution: x=3, y=p+1, z=1 x=p+1, y=3, z=1
$$\left({i}\right)−\left({ii}\right): \\ $$$${x}\left({y}−\mathrm{1}\right)+{z}^{\mathrm{2}} \left(\mathrm{1}−{y}\right)=\mathrm{2}{p} \\ $$$$\left({x}−{z}^{\mathrm{2}} \right)\left({y}−\mathrm{1}\right)=\mathrm{2}{p}=\begin{cases}{\mathrm{1}×\mathrm{2}{p}}\\{\mathrm{2}×{p}}\end{cases} \\ $$$$ \\ $$$${y}−\mathrm{1}=\mathrm{1},\:{x}−{z}^{\mathrm{2}} =\mathrm{2}{p} \\ $$$${y}=\mathrm{2},\:{x}={z}^{\mathrm{2}} +\mathrm{2}{p} \\ $$$${z}^{\mathrm{2}} +\mathrm{2}{p}+\mathrm{2}{z}^{\mathrm{2}} ={p}+\mathrm{4} \\ $$$$\mathrm{3}{z}^{\mathrm{2}} =\mathrm{4}−{p}\:\Rightarrow{not}\:{possible}, \\ $$$${since}\:{LHS}={odd}\:\geqslant\mathrm{3},\:{RHS}={even}\leqslant\mathrm{2} \\ $$$$ \\ $$$${y}−\mathrm{1}=\mathrm{2}{p},\:{x}−{z}^{\mathrm{2}} =\mathrm{1} \\ $$$${y}=\mathrm{2}{p}+\mathrm{1},\:{x}={z}^{\mathrm{2}} +\mathrm{1} \\ $$$${z}^{\mathrm{2}} +\mathrm{1}+\left(\mathrm{2}{p}+\mathrm{1}\right){z}^{\mathrm{2}} ={p}+\mathrm{4} \\ $$$$\mathrm{2}\left({p}+\mathrm{1}\right){z}^{\mathrm{2}} ={p}+\mathrm{3} \\ $$$$\left({p}+\mathrm{1}\right)\left(\mathrm{2}{z}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{2}\:\Rightarrow{not}\:{possible}, \\ $$$${since}\:{LHS}\geqslant\mathrm{3}\: \\ $$$$ \\ $$$${y}−\mathrm{1}=\mathrm{2},\:{x}−{z}^{\mathrm{2}} ={p} \\ $$$${y}=\mathrm{3},\:{x}={z}^{\mathrm{2}} +{p} \\ $$$${z}^{\mathrm{2}} +{p}+\mathrm{3}{z}^{\mathrm{2}} ={p}+\mathrm{4} \\ $$$${z}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{z}=\mathrm{1},\:{x}=\mathrm{1}+{p}\:\checkmark \\ $$$$ \\ $$$${y}−\mathrm{1}={p},\:{x}−{z}^{\mathrm{2}} =\mathrm{2} \\ $$$${y}={p}+\mathrm{1},\:{x}={z}^{\mathrm{2}} +\mathrm{2} \\ $$$${z}^{\mathrm{2}} +\mathrm{2}+\left({p}+\mathrm{1}\right){z}^{\mathrm{2}} ={p}+\mathrm{4} \\ $$$${z}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{z}=\mathrm{1},\:{x}=\mathrm{3},\:{y}={p}+\mathrm{1}\:\checkmark \\ $$$$ \\ $$$${summary}\:{of}\:{solution}: \\ $$$${x}=\mathrm{3},\:{y}={p}+\mathrm{1},\:{z}=\mathrm{1} \\ $$$${x}={p}+\mathrm{1},\:{y}=\mathrm{3},\:{z}=\mathrm{1} \\ $$
Commented by HongKing last updated on 05/Dec/21
very nice thank you dear Sir
$$\mathrm{very}\:\mathrm{nice}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Sir} \\ $$
Answered by Rasheed.Sindhi last updated on 06/Dec/21
{ ((xy + z^2 = 3p + 4)),((x + yz^2 = p + 4)) :} { ((x=((3p + 4−z^2 )/y))),((x= p + 4−yz^2 )) :} ⇒3p + 4−z^2 =py + 4y−y^2 z^2 y^2 z^2 −z^2 +3p−py+4−4y=0 z^2 (y−1)(y+1)−4(y−1)=p(y−3) determinant ((((y−1)( z^2 (y+1)−4 )=p(y−3)))) •∴ y−1=p ∧ z^2 (y+1)−4=y−3 y=p+1 ∧ z^2 (p+1+1)−4=p+1−3 z^2 =((p+2)/(p+2))=1⇒z=1 x=p+4−(p+1)(1)=3 (x,y,z)=(3,p+1,1) • (y−1)∣(y−3)⇒((y−3)/(y−1))∈{0,1,2,...} y−1=±1 or y−3=0 y=2,0,3 y=2 ⇒ z^2 (2+1)−4=p(2−3) z^2 =((−p+4)/3)⇒z=(√((4−p)/3)) ⇒p=1∉P y≠0 ∵ x=((3p + 4−z^2 )/y) y=3 ⇒ (3−1)( z^2 (3+1)−4 )=p(3−3) z^2 −1=0⇒z=1 x= p + 4−yz^2 =p+4−(3)(1)^2 x=p+1 (x,y,z)=(p+1,3,1)
$$\begin{cases}{\mathrm{xy}\:+\:\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{3p}\:+\:\mathrm{4}}\\{\mathrm{x}\:+\:\mathrm{yz}^{\mathrm{2}} \:=\:\mathrm{p}\:+\:\mathrm{4}}\end{cases}\: \\ $$$$\begin{cases}{\mathrm{x}=\frac{\mathrm{3p}\:+\:\mathrm{4}−\mathrm{z}^{\mathrm{2}} }{\mathrm{y}}}\\{\mathrm{x}=\:\mathrm{p}\:+\:\mathrm{4}−\mathrm{yz}^{\mathrm{2}} }\end{cases}\:\:\Rightarrow\mathrm{3p}\:+\:\mathrm{4}−\mathrm{z}^{\mathrm{2}} =\mathrm{py}\:+\:\mathrm{4y}−\mathrm{y}^{\mathrm{2}} \mathrm{z}^{\mathrm{2}} \\ $$$$\mathrm{y}^{\mathrm{2}} \mathrm{z}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} +\mathrm{3p}−\mathrm{py}+\mathrm{4}−\mathrm{4y}=\mathrm{0} \\ $$$$\mathrm{z}^{\mathrm{2}} \left(\mathrm{y}−\mathrm{1}\right)\left(\mathrm{y}+\mathrm{1}\right)−\mathrm{4}\left(\mathrm{y}−\mathrm{1}\right)=\mathrm{p}\left(\mathrm{y}−\mathrm{3}\right) \\ $$$$\begin{array}{|c|}{\left(\mathrm{y}−\mathrm{1}\right)\left(\:\mathrm{z}^{\mathrm{2}} \left(\mathrm{y}+\mathrm{1}\right)−\mathrm{4}\:\right)=\mathrm{p}\left(\mathrm{y}−\mathrm{3}\right)}\\\hline\end{array} \\ $$$$\bullet\therefore\:\mathrm{y}−\mathrm{1}=\mathrm{p}\:\wedge\:\mathrm{z}^{\mathrm{2}} \left(\mathrm{y}+\mathrm{1}\right)−\mathrm{4}=\mathrm{y}−\mathrm{3} \\ $$$$\underline{\mathrm{y}=\mathrm{p}+\mathrm{1}}\:\wedge\:\mathrm{z}^{\mathrm{2}} \left(\mathrm{p}+\mathrm{1}+\mathrm{1}\right)−\mathrm{4}=\mathrm{p}+\mathrm{1}−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{z}^{\mathrm{2}} =\frac{\mathrm{p}+\mathrm{2}}{\mathrm{p}+\mathrm{2}}=\mathrm{1}\Rightarrow\mathrm{z}=\mathrm{1} \\ $$$$\mathrm{x}=\mathrm{p}+\mathrm{4}−\left(\mathrm{p}+\mathrm{1}\right)\left(\mathrm{1}\right)=\mathrm{3} \\ $$$$\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{3},\mathrm{p}+\mathrm{1},\mathrm{1}\right) \\ $$$$\bullet\:\left(\mathrm{y}−\mathrm{1}\right)\mid\left(\mathrm{y}−\mathrm{3}\right)\Rightarrow\frac{\mathrm{y}−\mathrm{3}}{\mathrm{y}−\mathrm{1}}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…\right\} \\ $$$$\mathrm{y}−\mathrm{1}=\pm\mathrm{1}\:\mathrm{or}\:\mathrm{y}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{y}=\mathrm{2},\mathrm{0},\mathrm{3} \\ $$$$\:\:\:\:\mathrm{y}=\mathrm{2}\:\Rightarrow\:\mathrm{z}^{\mathrm{2}} \left(\mathrm{2}+\mathrm{1}\right)−\mathrm{4}=\mathrm{p}\left(\mathrm{2}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\mathrm{z}^{\mathrm{2}} =\frac{−\mathrm{p}+\mathrm{4}}{\mathrm{3}}\Rightarrow\mathrm{z}=\sqrt{\frac{\mathrm{4}−\mathrm{p}}{\mathrm{3}}}\:\Rightarrow\mathrm{p}=\mathrm{1}\notin\mathbb{P} \\ $$$$\:\:\:\:\:\mathrm{y}\neq\mathrm{0}\:\because\:\mathrm{x}=\frac{\mathrm{3p}\:+\:\mathrm{4}−\mathrm{z}^{\mathrm{2}} }{\mathrm{y}} \\ $$$$\underline{\mathrm{y}=\mathrm{3}}\:\Rightarrow\:\left(\mathrm{3}−\mathrm{1}\right)\left(\:\mathrm{z}^{\mathrm{2}} \left(\mathrm{3}+\mathrm{1}\right)−\mathrm{4}\:\right)=\mathrm{p}\left(\mathrm{3}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{z}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{z}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{x}=\:\mathrm{p}\:+\:\mathrm{4}−\mathrm{yz}^{\mathrm{2}} =\mathrm{p}+\mathrm{4}−\left(\mathrm{3}\right)\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{p}+\mathrm{1} \\ $$$$\:\:\:\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{p}+\mathrm{1},\mathrm{3},\mathrm{1}\right) \\ $$
Commented by mr W last updated on 06/Dec/21
just for discussion sir: in determinant ((((y−1)( z^2 (y+1)−4 )=p(y−3)))) we are not sure if y−3 is prime. if it is not prime, it can contain more factors, then there are much more possibilities.
$${just}\:{for}\:{discussion}\:{sir}: \\ $$$${in} \\ $$$$\begin{array}{|c|}{\left(\mathrm{y}−\mathrm{1}\right)\left(\:\mathrm{z}^{\mathrm{2}} \left(\mathrm{y}+\mathrm{1}\right)−\mathrm{4}\:\right)=\mathrm{p}\left(\mathrm{y}−\mathrm{3}\right)}\\\hline\end{array} \\ $$$${we}\:{are}\:{not}\:{sure}\:{if}\:{y}−\mathrm{3}\:{is}\:{prime}.\:{if} \\ $$$${it}\:{is}\:{not}\:{prime},\:{it}\:{can}\:{contain}\:{more} \\ $$$${factors},\:{then}\:{there}\:{are}\:{much}\:{more}\: \\ $$$${possibilities}. \\ $$
Commented by Rasheed.Sindhi last updated on 06/Dec/21
Yes sir I realized it.Now I′m going to change my answer in this light. Thanks to guide me sir!Please see my answer once more.
$${Yes}\:{sir}\:{I}\:{realized}\:{it}.{Now}\:{I}'{m}\:{going} \\ $$$${to}\:{change}\:{my}\:{answer}\:{in}\:{this}\:{light}. \\ $$$$\mathcal{T}{hanks}\:{to}\:{guide}\:{me}\:{sir}!{Please}\:{see} \\ $$$${my}\:{answer}\:{once}\:{more}. \\ $$
Commented by mr W last updated on 08/Dec/21
very good sir!
$${very}\:{good}\:{sir}! \\ $$
Commented by Rasheed.Sindhi last updated on 09/Dec/21
Grateful sir!
$${Grateful}\:\boldsymbol{{sir}}! \\ $$

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