x-2-y-3-z-4-have-integer-solutions-

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Question Number 425 by 123456 last updated on 25/Jan/15

$$x^2+y^3=z^4\mathrm{have}\mathrm{integer}\mathbf{solutions}?$$

Answered by prakash jain last updated on 02/Jan/15

$$\mathrm{Assuming}\mathrm{we}\mathrm{are}\mathrm{looking}\mathrm{for}+\mathrm{ve}$$$$\mathrm{solutions}$$$$y^3=(z^2-x)(z^2+x)$$$$\mathrm{To}\mathrm{find}\mathrm{a}\mathrm{solution}\mathrm{we}\mathrm{need}\mathrm{to}$$$$\mathrm{factorize}{y}^3=b\times a(b⩾a)\mathrm{such}\mathrm{that}$$$$a+\frac{b-a}{2}\mathrm{is}\mathrm{a}\mathrm{perfect}\mathrm{square}\mathrm{or}\frac{b+a}{2}$$$$\mathrm{is}\mathrm{perfect}\mathrm{square}.$$$$\mathrm{Then}z=\sqrt{\frac{b+a}{2}}$$$$x=\frac{b-a}{2}$$$$y=\sqrt[3]{b\times a}$$$$\mathrm{Trying}\mathrm{with}\mathrm{multiple}\mathrm{value}\mathrm{of}y,y=8$$$$\mathrm{gives}\mathrm{one}\mathrm{solution}$$$$y^3=8^3=8\times 64,a=8,b=64$$$$z=\sqrt{\frac{a+b}{2}}=6$$$$x=\frac{64-8}{2}=28$$$$x=28,y=8,z=6$$$$\mathrm{Also}\mathrm{if}x,y,z\mathrm{is}a\mathrm{solution}\mathrm{then}{n}^{6}x,n^{4}y,n^{3}z\mathrm{also}\mathrm{is}\mathrm{a}\mathrm{solution}.$$$$\mathbf{Othe}\mathbf{solution}\mathbf{with}\mathbf{higher}\mathbf{values}\mathbf{of}\mathit{y}.$$$$x=27,y=18,z=9y^3=54\times 108$$

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