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

- anonymous

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

- schrodinger

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- UnkleRhaukus

*

- anonymous

Let 8p+120 =n(n+1)/2 where n is a positive integer
16p+240 = n(n+1)

- anonymous

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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## More answers

- anonymous

Similarly we can do
p(16p +240/p)
2p(8p+120/p)
.
.
.

- anonymous

From there we get p can only be 1or 2 or 3 or 5 as 240 =2*2*2*2*3*5

- anonymous

But when p=1 or 3 or 5 we will have no solution
So, I got p=2 only

- anonymous

Did I miss anything?

- shubhamsrg

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.

- shubhamsrg

Can anyone tell me why 256 fails this ? It follows all conditions !

- anonymous

not prime

- shubhamsrg

n=256 = 32*8
256 *257/ 16 -15 =4097 = 17*241 => 2 different prime numbers.

- shubhamsrg

but 257 is prime, 256 should have worked ?

- shubhamsrg

352 also doesn't work, when simplified with n=352,we get 7751, which is 23*337, again a product of 2 primes !

- shubhamsrg

Something is fishy, with my hypothesis .

- shubhamsrg

exactly same is the case with 448 ! :|

- ParthKohli

@sauravshakya brilliant.org?!

- shubhamsrg

brilliant.org blocked me on fb! :P

- ParthKohli

And there are only two primes that satisfy this condition: \(2\) and \(47\).

- ParthKohli

Why, @shubhamsrg

- ParthKohli

I had written a program to solve this, oh well.

- shubhamsrg

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

- ParthKohli

You're such a genius heh :-)

- shubhamsrg

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|

- ParthKohli

This problem is from Brilliant.org by the way.
Have you applied for the summer camp?

- ParthKohli

@sauravshakya
@shubhamsrg
What are your profiles on there?

- shubhamsrg

Well I haven't gone on brilliant.org till now
I only liked it on fb .
Nothing else.

- ParthKohli

What will your age be in August?

- anonymous

I just made a account yesterday

- ParthKohli

Ah, that's bad. Students must be 13-17 for applying.

- ParthKohli

A free trip to Stanford. =)

- anonymous

I found some problem very interesting

- anonymous

That was the only reason I made account

- ParthKohli

I have done all problems except the last one.

- anonymous

@ParthKohli
did u get f(x)=-13+9x-3x^2+x^3
in the second last problem

- ParthKohli

No, I did a little bit of hit-and-trial. :-)

- ParthKohli

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 \(141\).

- ParthKohli

So I just counted the number of multiples of \(3\) between \(24\) and \(141\)... and voila!

- ParthKohli

Do you guys want the formalized solution anyway?

- ParthKohli

All the set elements are \(2\) modulo \(3\), and adding three numbers that are \(2\) modulo \(3\) gives us a number divisible by \(3\).

- ParthKohli

We can simply assert that all sums will be divisible by \(3\), where the maximum is \(141\) and the minimum is \(24\).

- ParthKohli

I could do better than that... but oh well.

- mathslover

hey I had submitted a solution for this question @sauravshakya to brilliant.org let us see what happens. by the way it is a great site.

- mathslover

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 : http://nrich.maths.org/790&part=solution ) 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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