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Question Number 184607 by mr W last updated on 09/Jan/23

solve { ((x^2 −xy+y^2 =16)),((y^2 −yz+z^2 =25)),((z^2 −zx+x^2 =49)) :}

$${solve} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{16}}\\{{y}^{\mathrm{2}} −{yz}+{z}^{\mathrm{2}} =\mathrm{25}}\\{{z}^{\mathrm{2}} −{zx}+{x}^{\mathrm{2}} =\mathrm{49}}\end{cases} \\ $$

Commented by Frix last updated on 09/Jan/23

Exact solution possible but not useful...

$$\mathrm{Exact}\:\mathrm{solution}\:\mathrm{possible}\:\mathrm{but}\:\mathrm{not}\:\mathrm{useful}... \\ $$

Commented by MJS_new last updated on 11/Jan/23

the system { ((x^2 +xy+y^2 =a^2 )),((y^2 +yz+z^2 =b^2 )),((z^2 +zx+x^2 =c^2 )) :} can be solved like this: 1. solve (2)−(3) for z 2. insert in (1) and (2) 3. let y=px and solve (1) for x^2 4. insert in (2) which leads to p^2 +αp+β=0 but the system { ((x^2 −xy+y^2 =a^2 )),((y^2 −yz+z^2 =b^2 )),((z^2 −zx+x^2 =c^2 )) :} leads to p^4 +αp^3 +βp^2 +γp+δ=0 which can only be exactly solved in some cases. with a=4, b=5, c=7 it′s not possible.

$$\mathrm{the}\:\mathrm{system}\:\begin{cases}{{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} ={a}^{\mathrm{2}} }\\{{y}^{\mathrm{2}} +{yz}+{z}^{\mathrm{2}} ={b}^{\mathrm{2}} }\\{{z}^{\mathrm{2}} +{zx}+{x}^{\mathrm{2}} ={c}^{\mathrm{2}} }\end{cases}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved} \\ $$$$\mathrm{like}\:\mathrm{this}: \\ $$$$\mathrm{1}.\:\mathrm{solve}\:\left(\mathrm{2}\right)−\left(\mathrm{3}\right)\:\mathrm{for}\:{z} \\ $$$$\mathrm{2}.\:\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{3}.\:\mathrm{let}\:{y}={px}\:\mathrm{and}\:\mathrm{solve}\:\left(\mathrm{1}\right)\:\mathrm{for}\:{x}^{\mathrm{2}} \\ $$$$\mathrm{4}.\:\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{2}\right)\:\mathrm{which}\:\mathrm{leads}\:\mathrm{to}\:{p}^{\mathrm{2}} +\alpha{p}+\beta=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{system}\:\begin{cases}{{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} ={a}^{\mathrm{2}} }\\{{y}^{\mathrm{2}} −{yz}+{z}^{\mathrm{2}} ={b}^{\mathrm{2}} }\\{{z}^{\mathrm{2}} −{zx}+{x}^{\mathrm{2}} ={c}^{\mathrm{2}} }\end{cases}\:\mathrm{leads}\:\mathrm{to} \\ $$$${p}^{\mathrm{4}} +\alpha{p}^{\mathrm{3}} +\beta{p}^{\mathrm{2}} +\gamma{p}+\delta=\mathrm{0}\:\mathrm{which}\:\mathrm{can}\:\mathrm{only}\:\mathrm{be} \\ $$$$\mathrm{exactly}\:\mathrm{solved}\:\mathrm{in}\:\mathrm{some}\:\mathrm{cases}. \\ $$$$\mathrm{with}\:{a}=\mathrm{4},\:{b}=\mathrm{5},\:{c}=\mathrm{7}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{possible}. \\ $$

Commented by mr W last updated on 11/Jan/23

thanks sir!

$${thanks}\:{sir}! \\ $$

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