Are you smarter than me? 49 A
m ( 4 m 2 + m + 1 2 ) = 3 ( p n − 1 )
Find sum of all the positive integers m , n and p , where p is a prime greater than 4, such that they satisfy the equation above.
The answer is 23.
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1 solution
Thanks! I didn't interpret it as sum of m, n and p. LOL
about the same solution, but in the middle, i speculate the gcd of 12k+1 and 3k^2+1 is 7 or 49, so p=7. Then I used 3k^2+1has to be a multiple of 12k+1 and use ∆=b^2-4ac has to be a perfect square to solve
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Do you know how to prove your speculation? Where did 7 or 49 come from?
Hint: What is ( 1 2 k + 1 ) ( 1 2 k − 1 ) ?
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gcd(12k+1, 3k^2+1)=gcd(12k+1, (12k^2+k)-k+4) =gcd(12k+1,k-4) =gcd(12k+1, (12k+1)-49) =gcd(12k+1,49)=1 or 7 or 49. If the gcd is one, then either 12k+1 is one or 3k^2+1 is one (impossible). Thus, the gcd is 7 or 49. So p=7 anyway.
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@Zhiwei William Zhang – Right! Note that in the first equality is justified because both numbers are odd. Otherwise, my preferred approach would be to say that each number divides the next. IE
g cd ( 1 2 k + 1 , 3 k 2 + 1 ) ∣ g cd ( 1 2 k + 1 , 1 4 4 k 2 + 4 8 ) = g cd ( 1 2 k + 1 , ( 1 4 4 k 2 − 1 ) + 4 9 = g cd ( 1 2 k + 1 , 4 9 ) ∣ 4 9 .
The first ∣ is actually an equality sign, because g cd ( 1 2 k + 1 , 4 8 ) = 1 .
Rewrite the equation as m ( 4 m 2 + m + 1 2 ) + 3 = 3 p n ( 4 m + 1 ) ( m 2 + 3 ) = 3 p n Now LHS must be divisible by 3 , It follows that m ≡ 3 0 , 2
Case 1) m = 3 k ; k > 0 ( 1 2 k + 1 ) ( 3 k 2 + 1 ) = p n Examine some small values of k . k = 1 , 2 , 3 fail to give a sloution, k = 4 gives 7 4 = p n , therefore ( m , p , n ) = ( 1 2 , 7 , 4 ) is an answer. For k > 4 factors 1 2 k + 1 and 3 k 2 + 1 are greater than 1 and 3 k 2 + 1 > 1 2 k + 1 , hence both of them are some exponents of p and power of p in 3 k 2 + 1 is greater than 3 k + 1 , It's immediate that we must have 1 2 k + 1 ∣ 3 k 2 + 1 and 1 2 k + 1 ∣ k ( 1 2 k + 1 ) − 4 ( 3 k 2 + 1 ) = k − 4 1 2 k + 1 ∣ 1 2 k + 1 − 1 2 ( k − 4 ) = 4 9 1 2 k + 1 ∣ 4 9 Which is impossible for k > 4 .
Case 2) m = 3 k + 2 ; k ≥ 0 ( 4 k + 3 ) ( 9 k 2 + 1 2 k + 4 ) = p n Here 9 k 2 + 1 2 k + 4 > 4 k + 3 for all k ≥ 0 and with a similar reasoning to case one we must have 4 k + 3 ∣ 9 k 2 + 1 2 k + 4 4 k + 3 ∣ 4 ( 9 k 2 + 1 2 k + 4 ) − 9 k ( 4 k + 3 ) = 2 1 k + 1 6 4 k + 3 ∣ 4 ( 2 1 k + 1 6 ) − 2 1 ( 4 k + 3 ) = 1 4 k + 3 ∣ 1 This is impossible.
There is only one solution: ( m , p , n ) = ( 1 2 , 7 , 4 ) .