Try this beautiful problem from the PRMO, 2019 based on Smallest Positive Integer.
Find the smallest positive integer n\(\geq\)10 such that n+6 is a prime and 9n+7 is a perfect square.
Integers
Primes
Perfect Square
But try the problem first...
Answer: is 53.
PRMO, 2019, Question 14
Elementary Number Theory by David Burton
First hint
Let 9n+7=\(m^{2}\) n+6 prime then n+6 odd then n is odd then n=2k+1 then 9(2k+1)+7=\(m^{2}\) then 18k=\(m^{2}\)-16=(m+4)(m-4) then 18k even m is even then m=2p
Second Hint
18k=(2p+4)(2p-4)=4(p+2)(p-2) then 9k=2(p+2)(p-2)then k even then k=2d then 18d=2(p+2)(p-2) then 9d=(p+2)(p-2) then p of form 9q+2,9q-2
Final Step
for p=9q-2 then m=2(9q-2) for q=1 then\(m^{2}\)=196then n=21 then n+6=27 non prime, for p=9q+2 then m=2(9q+2) for q=1 \(m^{2}\)=484 then n=53 then n+6=59 prime then n=53.
Try this beautiful problem from the PRMO, 2019 based on Smallest Positive Integer.
Find the smallest positive integer n\(\geq\)10 such that n+6 is a prime and 9n+7 is a perfect square.
Integers
Primes
Perfect Square
But try the problem first...
Answer: is 53.
PRMO, 2019, Question 14
Elementary Number Theory by David Burton
First hint
Let 9n+7=\(m^{2}\) n+6 prime then n+6 odd then n is odd then n=2k+1 then 9(2k+1)+7=\(m^{2}\) then 18k=\(m^{2}\)-16=(m+4)(m-4) then 18k even m is even then m=2p
Second Hint
18k=(2p+4)(2p-4)=4(p+2)(p-2) then 9k=2(p+2)(p-2)then k even then k=2d then 18d=2(p+2)(p-2) then 9d=(p+2)(p-2) then p of form 9q+2,9q-2
Final Step
for p=9q-2 then m=2(9q-2) for q=1 then\(m^{2}\)=196then n=21 then n+6=27 non prime, for p=9q+2 then m=2(9q+2) for q=1 \(m^{2}\)=484 then n=53 then n+6=59 prime then n=53.
