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Find all prime numbers "p", such that 8p+120 is a triangle number? I got only 2

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Let 8p+120 =n(n+1)/2 where n is a positive integer 16p+240 = n(n+1)
Now, 2(8p+120) from here no solution as 2+1<120 4(4p+60) from here no solution as 4+1<60 8(2p+30) from here no solution as 8+1<30 16(p+15) here we have a solution when p=2 an i.e 16*17

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Similarly we can do p(16p +240/p) 2p(8p+120/p) . . .
From there we get p can only be 1or 2 or 3 or 5 as 240 =2*2*2*2*3*5
But when p=1 or 3 or 5 we will have no solution So, I got p=2 only
Did I miss anything?
16p + 240 = n(n+1) p = n(n+1)/16 - 15 = n(n+1)/16 - (3*5) n(n+1) should be divisible by 16, it should not be odd, also , it should not have 3 or 5 as factor. So, n or n+1 should be prime, one is prime then the other should not be divisible by 3 or 5,n(n+1) should be divisible by 16, n(n+1)/16 should be even. => n(n+1) should be divisible by 32 , given 2 is excluded as our prime. n=16 fits 31 fits I see no other number till now, p=2 and p=47 seems legit at the moment.
Can anyone tell me why 256 fails this ? It follows all conditions !
not prime
n=256 = 32*8 256 *257/ 16 -15 =4097 = 17*241 => 2 different prime numbers.
but 257 is prime, 256 should have worked ?
352 also doesn't work, when simplified with n=352,we get 7751, which is 23*337, again a product of 2 primes !
Something is fishy, with my hypothesis .
exactly same is the case with 448 ! :|
@sauravshakya! blocked me on fb! :P
And there are only two primes that satisfy this condition: \(2\) and \(47\).
I had written a program to solve this, oh well.
They had asked something, I posted a very neat and good solution in the comments, explaining it pretty well. After 2-3 days, when I checked back, They had deleted my comment! :O I abused them! :P And now the consequences! :P
You're such a genius heh :-)
Hell no! It was one of those rare moments when I was able to do some question. Hence I got too carried away when I came to know it wasn't there. B|
This problem is from by the way. Have you applied for the summer camp?
@sauravshakya @shubhamsrg What are your profiles on there?
Well I haven't gone on till now I only liked it on fb . Nothing else.
What will your age be in August?
I just made a account yesterday
Ah, that's bad. Students must be 13-17 for applying.
A free trip to Stanford. =)
I found some problem very interesting
That was the only reason I made account
I have done all problems except the last one.
@ParthKohli did u get f(x)=-13+9x-3x^2+x^3 in the second last problem
No, I did a little bit of hit-and-trial. :-)
The maximum number that you could get from the set was \(44 +47 + 50 =141\) and the least was \(24\). Then I just randomly added some set elements and I observed that you always get a multiple of \(3\) between \(24\) and \(141\).
So I just counted the number of multiples of \(3\) between \(24\) and \(141\)... and voila!
Do you guys want the formalized solution anyway?
All the set elements are \(2\) modulo \(3\), and adding three numbers that are \(2\) modulo \(3\) gives us a number divisible by \(3\).
We can simply assert that all sums will be divisible by \(3\), where the maximum is \(141\) and the minimum is \(24\).
I could do better than that... but oh well.
hey I had submitted a solution for this question @sauravshakya to let us see what happens. by the way it is a great site.
Go for hit and trial method first. We have the first prime number as 2 : let us take p = 2 and see what we get : 8p + 120 = 8(2) + 120 = 16+120 = 136. Is 136 in the form of n(n+1)/2 ? Let us check it . n(n+1)/2 = 136 n(n+1) = 272 n^2 + n - 272 = 0 from this we get n = 16 and so we get : 16 * 17/2 = 136. That is 2 is our first prime number that satisfies the given equation : 8p + 120 . See, we know that a triangular number forms an equilateral triangle . i.e. 8p + 120 will form an equilateral triangle. We know that if n is a triangular number then 8n + 1 is a square ( find the proof here : ) therefore we have : 8(8p + 120) + 1 is a square 64 p + 961 will be a square. Let the side of that square be a therefore we have : 64 p + 961 = a^2 a^2 - 961 = 64 p a^2 - 31^2 = 64p (a+31)(a-31) = 64p . We have : (a+31)(a-31) = 64p Let us first take a as odd Since a is odd therefore it must be in the form of 2k+1 ( k is even ) Therefore we have : (2k + 32)(2k – 30) = 64p (k + 16)(k – 15) = 16p Since k is even therefore it can be written in the form of \(2\lambda\) \((2\lambda + 16)(2\lambda – 15) = 16p\) \((\lambda+8) (2\lambda – 15) = 8p\) Therefore \(8|\lambda +8\) since 8 divides 8 and so it must divide also Therefore \(8| \lambda\) or \(16|2\lambda\) or \(16|k\) (We have taken earlier as k = \(2\lambda\)) Since 16 divides k so we can write k as \(16\alpha\) . Putting this in the equation : (2k+32)(2k-30) = 64p \((2(16\alpha)+32)(2(16\alpha)-30) = 64p\) \((32\alpha+64)(32\alpha -30) = 64p\) \((\alpha+1)(16\alpha-15)=p\) Oh! Here we have written a prime number ( p ) in the form of two composite numbers multiplication. But we have done everything right so there must be like : \(16\alpha – 15 = 1\) and \(\alpha + 1 = p\) Therefore \(alpha = 1\) and p = 2 which we have already found. This means k is not in the form of \(2\lambda\) Let us take it now as \(k = 2\delta + 1\) We have : \((k+16)(k-15) = 16p\) - this we got earlier \((2\delta+1 + 16)(2\delta +1 – 15) = 16p\) \((2\delta+17)(2\delta – 14) = 16p\) \((2\delta+17)(\delta – 7) = 8p\) Since \(2\delta + 17\) is odd ( since \(2\delta\) is even and even + odd = odd ) therefore 8 must divide \(\delta-7\) . Let us take \(\delta – 7 \) = \(8 \beta \) \(\delta = 8\beta + 7\) Since k = \(2\delta + 1\) therefore we have : \(k = 2(8\beta +7)+1\) and thus we get : \((k+16)(k-15) = 16p\) as : \((16\beta + 31)(16\beta) = 16p\) \((16\beta + 31)(\beta) = p\) Again we have p as the multiplication of two composite numbers . So it must be like \(\beta= 1\) and \(16\beta+31 = p\) Therefore \(\beta = 1\) and \(16\beta + 31 = 16(1) + 31 = 16+31 = 47 = p \) Therefore another prime number is 47 and so the sum of the prime numbers is 49 . ---**************************************---------------------------------------------------*****************************************

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