[آكل الداء]

x^2-px+q=0
(x-a1)(x-a2)=0
Then q=a1a2,p=a1+a2
as q>0 then a1,a2>0
Then a1=1,a2=q
p=q+1
q=2,p=3
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The numbers 2*3*5*7*11*13*17*19+1+i
i denote the order of the number between the 20 numbers
prove is not very hard take commen factors
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x^2+ax+b+1=0
(x-x1)(x-x2)=0
a=-(x1+x2)
b=x1x2-1
a^2+b^2=(x1^2+1)(x2^2+1)
both of brackets graeter then or equal 2 then not prime
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case1:n even=>obvious
case2:n odd=>n=2k+1
n^4+4^n=(n^4)+4(2^k)^4
we have Sophie Formula a^4+4(b^4)=(a^2+2b^2+2(ab)^2)(a^2+b^2-2(ab)^2)
then n^4+4^n=(n^2+2^2k+1+2(n^2*2^k))(n^2+2^2k+1-2(n^2*2^k))
first bracket positive if the second =1 then (1-n^2)(2^n-1)=0
either n=1 or 0 which is not possible
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1/x(x+1)=1/x - 1/x+1
then 1- 1/n+1=R.H.S
n/n+1=16+n+(4-n)^1/2 /21+n
then (4-n)^1/2 real number n is integer the 2 possible cases n=3 or n=4
by examing only solution n=3