Menu Close

Show-that-if-3-prime-numbers-all-greater-than-3-form-an-arithmetic-progression-then-the-common-difference-of-the-progression-is-divisible-by-6-

Tinku Tara 0 Comments
Pin




Question Number 92673 by Ar Brandon last updated on 08/May/20
Show that if 3 prime numbers, all greater than 3, form an arithmetic progression then the common difference of the progression is divisible by 6.
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{3}\:\mathrm{prime}\:\mathrm{numbers},\:\mathrm{all}\:\mathrm{greater} \\ $$$$\mathrm{than}\:\mathrm{3},\:\mathrm{form}\:\mathrm{an}\:\mathrm{arithmetic}\:\mathrm{progression}\:\mathrm{then}\:\mathrm{the}\:\mathrm{common} \\ $$$$\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{progression}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{6}. \\ $$
Commented by Rasheed.Sindhi last updated on 09/May/20
In other words: If a & b are two primes such that 6 ∤ (b−a) then 2b−a is composite.
$${In}\:{other}\:{words}: \\ $$$${If}\:\:{a}\:\:\&\:\:{b}\:{are}\:{two}\:{primes}\:{such}\:{that} \\ $$$$\mathrm{6}\:\nmid\:\left({b}−{a}\right)\:{then}\:\mathrm{2}{b}−{a}\:{is}\:{composite}. \\ $$
Commented by Ar Brandon last updated on 09/May/20
Interesting Sir. I'll like to get more details please.
Answered by Rasheed.Sindhi last updated on 10/May/20
Let a,b∈P are first two terms ∴ b−a is common difference ∴ The Third is b+(b−a)=2b−a According to the given if (2b−a)∈P with a,b∈P, then 6∣(b−a). In other words if 6 ∤ (b−a) then 2b−a is not prime i-e composite. Proof: ^• All integers (w. r. t division by 6) can be categarized into six types: 6k,6k+1,6k+2,...,6k+5. Among these 6k,6k+2,6k+3 & 6k+4 are obviously composite. This means primes are only of two types: 6k+1 & 6k+5 Possibility for prime pair (a,b) are (6k+1,6k+1) , (6k+1,6k+5) (6k+5,6k+1) & (6k+5,6k+5) and b−a will respectively befall in types:6k,6k+2,6k+4 &6k. That means the pairs for which a,b∈P and 6 ∤ (b−a) (6k+1,6k+5) & (6k+5,6k+1) AND we′ve to prove for these pairs that 2b−a(the third term) is not prime i-e composite. see the table determinant ((a,b,(b−a),(2b−a)),((6k+1),(6k+1),(6k_(6∣(b−a)) ),(6k+1)),((6k+1),(6k+5),(6k+4_(6 ∤ (b−a)) ),(6k+3^■ _(NonPrime) )),((6k+5),(6k+1),(6k+2_(6 ∤ (b−a)) ),(6k+3^■ _(NonPrime) )),((6k+5),(6k+5),(6k_(6∣(b−a)) ),(6k+5))) ^■ 6k+3 is nonprime for k>0 that is for a,b>3 Red rows represent that if b−a (common difference) can′t be divided by 6, the third term can′t be prime. That is if three prime numbers(>3) are in AP ,then 6 must divide common difference. Q.E.D
$${Let}\:{a},{b}\in\mathbb{P}\:{are}\:{first}\:{two}\:{terms} \\ $$$$\therefore\:{b}−{a}\:{is}\:{common}\:{difference} \\ $$$$\therefore\:{The}\:{Third}\:{is}\:{b}+\left({b}−{a}\right)=\mathrm{2}{b}−{a} \\ $$$${According}\:{to}\:{the}\:{given}\:{if}\:\left(\mathrm{2}{b}−{a}\right)\in\mathbb{P} \\ $$$${with}\:{a},{b}\in\mathbb{P},\:{then}\:\mathrm{6}\mid\left({b}−{a}\right). \\ $$$${In}\:{other}\:{words}\:{if}\:\mathrm{6}\:\nmid\:\left({b}−{a}\right)\:{then} \\ $$$$\mathrm{2}{b}−{a}\:{is}\:{not}\:{prime}\:{i}-{e}\:{composite}. \\ $$$${Proof}: \\ $$$$\:^{\bullet} {All}\:{integers}\:\left({w}.\:{r}.\:{t}\:{division}\:{by}\:\mathrm{6}\right) \\ $$$$\:{can}\:{be}\:{categarized}\:{into}\:{six}\:{types}: \\ $$$$\:\mathrm{6}{k},\mathrm{6}{k}+\mathrm{1},\mathrm{6}{k}+\mathrm{2},…,\mathrm{6}{k}+\mathrm{5}. \\ $$$$\:\:{Among}\:{these}\:\mathrm{6}{k},\mathrm{6}{k}+\mathrm{2},\mathrm{6}{k}+\mathrm{3}\:\& \\ $$$$\:\:\mathrm{6}{k}+\mathrm{4}\:{are}\:{obviously}\:{composite}. \\ $$$$\:\:{This}\:{means}\:{primes}\:{are}\:{only}\:{of} \\ $$$$\:{two}\:{types}:\:\mathrm{6}{k}+\mathrm{1}\:\&\:\mathrm{6}{k}+\mathrm{5} \\ $$$$\:{Possibility}\:{for}\:{prime}\:{pair}\:\left({a},{b}\right) \\ $$$$\:\:{are}\:\left(\mathrm{6}{k}+\mathrm{1},\mathrm{6}{k}+\mathrm{1}\right)\:,\:\left(\mathrm{6}{k}+\mathrm{1},\mathrm{6}{k}+\mathrm{5}\right) \\ $$$$\:\:\left(\mathrm{6}{k}+\mathrm{5},\mathrm{6}{k}+\mathrm{1}\right)\:\&\:\left(\mathrm{6}{k}+\mathrm{5},\mathrm{6}{k}+\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{and} \\ $$$$\:\:{b}−{a}\:{will}\:\:{respectively}\:{befall}\:{in} \\ $$$$\:{types}:\mathrm{6}{k},\mathrm{6}{k}+\mathrm{2},\mathrm{6}{k}+\mathrm{4}\:\&\mathrm{6}{k}. \\ $$$$\:{That}\:{means}\:{the}\:{pairs}\:{for}\:{which}\: \\ $$$$\:\:{a},{b}\in\mathbb{P}\:{and}\:\mathrm{6}\:\nmid\:\left({b}−{a}\right)\: \\ $$$$\:\:\left(\mathrm{6}{k}+\mathrm{1},\mathrm{6}{k}+\mathrm{5}\right)\:\&\:\left(\mathrm{6}{k}+\mathrm{5},\mathrm{6}{k}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{AND} \\ $$$$\:\:\:{we}'{ve}\:{to}\:{prove}\:{for}\:{these}\:{pairs} \\ $$$$\:\:\:{that}\:\mathrm{2}{b}−{a}\left({the}\:{third}\:{term}\right)\:{is}\: \\ $$$$\:\:{not}\:{prime}\:{i}-{e}\:{composite}.\:{see} \\ $$$$\:\:\:{the}\:{table} \\ $$$$\:\begin{vmatrix}{{a}}&{{b}}&{{b}−{a}}&{\mathrm{2}{b}−{a}}\\{\mathrm{6}{k}+\mathrm{1}}&{\mathrm{6}{k}+\mathrm{1}}&{\underset{\mathrm{6}\mid\left({b}−{a}\right)} {\mathrm{6}{k}}}&{\mathrm{6}{k}+\mathrm{1}}\\{\mathrm{6}{k}+\mathrm{1}}&{\mathrm{6}{k}+\mathrm{5}}&{\underset{\mathrm{6}\:\nmid\:\left({b}−{a}\right)} {\mathrm{6}{k}+\mathrm{4}}}&{\underset{{NonPrime}} {\mathrm{6}{k}+\mathrm{3}^{\blacksquare} }}\\{\mathrm{6}{k}+\mathrm{5}}&{\mathrm{6}{k}+\mathrm{1}}&{\underset{\mathrm{6}\:\nmid\:\left({b}−{a}\right)} {\mathrm{6}{k}+\mathrm{2}}}&{\underset{{NonPrime}} {\mathrm{6}{k}+\mathrm{3}^{\blacksquare} }}\\{\mathrm{6}{k}+\mathrm{5}}&{\mathrm{6}{k}+\mathrm{5}}&{\underset{\mathrm{6}\mid\left({b}−{a}\right)} {\mathrm{6}{k}}}&{\mathrm{6}{k}+\mathrm{5}}\end{vmatrix} \\ $$$$\:^{\blacksquare} \:\mathrm{6}{k}+\mathrm{3}\:{is}\:{nonprime}\:{for}\:{k}>\mathrm{0} \\ $$$$\:\:\:\:\:\:{that}\:{is}\:{for}\:{a},{b}>\mathrm{3} \\ $$$${Red}\:{rows}\:{represent}\:{that}\:{if}\:{b}−{a} \\ $$$$\left({common}\:{difference}\right)\:{can}'{t} \\ $$$${be}\:{divided}\:{by}\:\mathrm{6},\:{the}\:{third}\:{term} \\ $$$${can}'{t}\:{be}\:{prime}. \\ $$$${That}\:{is}\:{if}\:{three}\:{prime}\:{numbers}\left(>\mathrm{3}\right) \\ $$$${are}\:{in}\:{AP}\:,{then}\:\mathrm{6}\:{must}\:{divide}\:{common} \\ $$$${difference}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Q}.\mathcal{E}.{D} \\ $$
Commented by Ar Brandon last updated on 10/May/20
Thank you ��

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *