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Suppose-an-integer-x-a-natural-number-n-and-a-prime-number-p-satisfy-the-equation-7x-2-44x-12-p-n-Find-the-largest-value-of-p-

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Question Number 21200 by Tinkutara last updated on 15/Sep/17
Suppose an integer x, a natural number n and a prime number p satisfy the equation 7x^2 − 44x + 12 = p^n . Find the largest value of p.
$$\mathrm{Suppose}\:\mathrm{an}\:\mathrm{integer}\:{x},\:\mathrm{a}\:\mathrm{natural} \\ $$$$\mathrm{number}\:{n}\:\mathrm{and}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:{p} \\ $$$$\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{7}{x}^{\mathrm{2}} \:−\:\mathrm{44}{x}\:+\:\mathrm{12}\:=\:{p}^{{n}} . \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{value}\:\mathrm{of}\:{p}. \\ $$
Commented by mrW1 last updated on 16/Sep/17
47?
$$\mathrm{47}? \\ $$
Answered by alex041103 last updated on 16/Sep/17
7x^2 −44x+12=(x−6)(7x−2)=p^n ⇒x−6=p^n_1 and 7x−2=p^n_2 and n_1 +n_2 =n ⇒x=p^n_1 +6 and 7p^n_1 +40=p^n_2 ⇒n_1 <n_2 ⇒7+((40)/p^n_1 ) is a whole number ⇒p^n_1 is a divisor of 40=2^3 ×5^1 ×(som num)^0 (p,n_1 )=(p,0),(5,1),(2,3) Now we try p=5 or 2 and it doesn′t work ⇒p^n_2 =7×p^0 +40=47 (47 is prime) ⇒p=47, n_1 =0, n_2 =1, n=1 Ans. p_(max) =p=47 and 7x^2 −44x+12= =7×7^2 −44×7 +12= 47^1 =p^n
$$\mathrm{7}{x}^{\mathrm{2}} −\mathrm{44}{x}+\mathrm{12}=\left({x}−\mathrm{6}\right)\left(\mathrm{7}{x}−\mathrm{2}\right)={p}^{{n}} \\ $$$$\Rightarrow{x}−\mathrm{6}={p}^{{n}_{\mathrm{1}} } \:{and}\:\mathrm{7}{x}−\mathrm{2}={p}^{{n}_{\mathrm{2}} } \:{and}\:{n}_{\mathrm{1}} +{n}_{\mathrm{2}} ={n} \\ $$$$\Rightarrow{x}={p}^{{n}_{\mathrm{1}} } +\mathrm{6}\:{and}\:\mathrm{7}{p}^{{n}_{\mathrm{1}} } +\mathrm{40}={p}^{{n}_{\mathrm{2}} } \\ $$$$\Rightarrow{n}_{\mathrm{1}} <{n}_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{7}+\frac{\mathrm{40}}{{p}^{{n}_{\mathrm{1}} } }\:{is}\:{a}\:{whole}\:{number} \\ $$$$\Rightarrow{p}^{{n}_{\mathrm{1}} } \:{is}\:{a}\:{divisor}\:{of}\:\mathrm{40}=\mathrm{2}^{\mathrm{3}} ×\mathrm{5}^{\mathrm{1}} ×\left({som}\:{num}\right)^{\mathrm{0}} \\ $$$$\left({p},{n}_{\mathrm{1}} \right)=\left({p},\mathrm{0}\right),\left(\mathrm{5},\mathrm{1}\right),\left(\mathrm{2},\mathrm{3}\right) \\ $$$${Now}\:{we}\:{try}\:{p}=\mathrm{5}\:{or}\:\mathrm{2}\:{and}\:{it}\:{doesn}'{t}\:{work} \\ $$$$\Rightarrow{p}^{{n}_{\mathrm{2}} } =\mathrm{7}×{p}^{\mathrm{0}} +\mathrm{40}=\mathrm{47}\:\left(\mathrm{47}\:{is}\:{prime}\right) \\ $$$$\Rightarrow{p}=\mathrm{47},\:{n}_{\mathrm{1}} =\mathrm{0},\:{n}_{\mathrm{2}} =\mathrm{1},\:{n}=\mathrm{1} \\ $$$${Ans}.\:{p}_{{max}} ={p}=\mathrm{47}\:{and}\: \\ $$$$\mathrm{7}{x}^{\mathrm{2}} −\mathrm{44}{x}+\mathrm{12}= \\ $$$$=\mathrm{7}×\mathrm{7}^{\mathrm{2}} −\mathrm{44}×\mathrm{7}\:+\mathrm{12}=\:\mathrm{47}\:^{\mathrm{1}} ={p}^{{n}} \\ $$$$ \\ $$$$ \\ $$
Commented by Tinkutara last updated on 16/Sep/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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