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Question Number 21200 by Tinkutara last updated on 15/Sep/17

$$\mathrm{Suppose}\mathrm{an}\mathrm{integer}x,\mathrm{a}\mathrm{natural}$$$$\mathrm{number}n\mathrm{and}\mathrm{a}\mathrm{prime}\mathrm{number}p$$$$\mathrm{satisfy}\mathrm{the}\mathrm{equation}7x^2-44x+12=p^n.$$$$\mathrm{Find}\mathrm{the}\mathrm{largest}\mathrm{value}\mathrm{of}p.$$

Commented by mrW1 last updated on 16/Sep/17

$$47?$$

Answered by alex041103 last updated on 16/Sep/17

$$7x^2-44x+12=(x-6)(7x-2)=p^n$$$$\Rightarrow x-6=p^{n_1}and7x-2=p^{n_2}and{n}_1+n_2=n$$$$\Rightarrow x=p^{n_1}+6and7p^{n_1}+40=p^{n_2}$$$$\Rightarrow n_1<n_2$$$$\Rightarrow 7+\frac{40}{p^{n_1}}isawholenumber$$$$\Rightarrow p^{n_1}isadivisorof40=2^3\times 5^1\times {\left(somnum\right)}^0$$$$(p,n_1)=(p,0),(5,1),(2,3)$$$$Nowwetryp=5or2andit{doesn}^{\prime }twork$$$$\Rightarrow p^{n_2}=7\times p^0+40=47\left(47isprime\right)$$$$\Rightarrow p=47,n_1=0,n_2=1,n=1$$$$Ans.p_{max}=p=47and$$$$7x^2-44x+12=$$$$=7\times 7^2-44\times 7+12=47{}^1=p^n$$$$$$$$$$

Commented by Tinkutara last updated on 16/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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