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Question Number 174512 by ahdal last updated on 03/Aug/22

$$howmanyintegera,b\in z^{+}$$$$a^5-b^5=10{(b+1)}^2-9$$$$$$

Answered by MJS_new last updated on 03/Aug/22

$$a^5=b^5+10b^2+20b+1$$$$b>0\Rightarrow a>b\Rightarrow a=b+n\wedge n\in {\mathbb{N}}^{★}$$$${(b+n)}^5=b^5+10b^2+20b+1$$$$5b^{4}n+10b^3{n}^2+10b^2{n}^3+5{bn}^4+n^5=10b^2+20b+1$$$$\mathrm{since}b⩾1\wedge n⩾1{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{obvious}\mathrm{that}\mathrm{the}\mathrm{lhs}\mathrm{is}$$$$\mathrm{greater}\mathrm{than}\mathrm{the}\mathrm{rhs}\mathrm{for}\mathrm{higher}\mathrm{values}\mathrm{of}$$$$\mathrm{both}b\mathrm{or}n.\mathrm{the}\mathrm{only}\mathrm{solution}\mathrm{is}$$$$b=1\wedge n=1\Rightarrow a=2$$

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