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Question Number 194444 by Abdullahrussell last updated on 06/Jul/23

Answered by Frix last updated on 08/Jul/23

$$\mathrm{Let}y=px\wedge z=qx$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{get}$$$$p=\frac{b(4-b)}{ab-2(a+b+c-4)}$$$$q=\frac{bc-2(a+b+c-4)}{ab-2(a+b+c-4)}$$$$ab-2(a+b+c-4)\ne 0$$$$\mathrm{But}\mathrm{we}\mathrm{are}\mathrm{now}\mathrm{free}\mathrm{to}\mathrm{choose}x$$$$\mathrm{If}\mathrm{we}\mathrm{first}\mathrm{choose}(x,y,z)\mathrm{to}\mathrm{get}(a,b,c)$$$$\mathrm{we}\mathrm{find}\mathrm{that}\mathrm{all}\mathrm{triples}(r,pr,qr)\mathrm{solve}\mathrm{the}$$$$\mathrm{given}\mathrm{equations}.$$

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