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Q1. Second principle of induction: prove that for all natural numbers n (3 + sqrt(5))^n + (3 - sqrt(5))^n is an even integer. Let S_n = (3 + sqrt(5))^n + (3 - sqrt(5))^n a = 3 + rt(5) b = 3 - rt(5) a+b = 6 ab = 4 a,b are roots of x^2 - 6x + 4 = 0 => x^2 = 6x - 4 Multiply both sides with x^(n-1) => x^(n+1) = 6x^(n) - 4x^(n-1) replace x with a,b and add them together S_{n+1} = 6S_{n} - 4S_{n-1} For n=1,2 Sn is even. Using second principle of induction assume it's true till S_{k} including S_{k-1} and below. Then S_{k+1} = 6S_{k} - 4S_{k-1}is also even. H.P.
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practice problems pending

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 Q1. S1. Method 1: sum = 0 i = 1, j = 1:1, k = 1:1 Add 1 i = 2, j = 1, k = 1:1 Add 1 i = 2, j = 2, k = 1:2 Add 2 First iteration we added: 1 Second:  (1+2) Third:  (1+2+3) Nth: . (1+2+3...+n) Adding all we get: 1 + (1+2) + (1+2+3) .... + (1+...n) = 1*n + 2*(n-1) .... 1*n = (k=1:n)Sigma (k*(n-k+1))  = sigma(nk - k^2 + k) = (n+1)*[n(n+1)]/2 - n(n+1)(2n+1)/6 = n(n+1)(n+2)/6 Method 2:  reduce it step by step: Triple sigma 1 = double sigma j = sigma i*(i+1)/2 then separate and do the sum. Q2. Find the sum of the products of every pair of the first n natural numbers. S2. 1*2 + 1*3 + 1*4 ... 1*n           2*3 + 2*4 ... 2*n ............................                                (n-1)*n = (k = n:2)Sigma(k * {k*(k-1)/2} = 1/2[sigma(n^3) - 1 - sigma(n^2) + 1] = 1/2[sigma(n^3) - sigma(n^2)] simplify from here and get the answer. Q3. Row 1: 1 Row 2:  2 3 Row 3:...
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Diophantine m-tuple by Andrej Dujella

 for a,b positive integers if ab+ 1 = k^2 i.e. perfect square then if we use c = a + b + 2k then bc + 1 and ac + 1 will also be perfect squares.
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Pell's equations

 x^2 - Dy^2 = 1 has infinitely many integer solutions (x,y) if D is a positive non-square integer. Trivial solution: (1,0) Fundamental solution: smallest positive values of x,y both of which are non-zero.
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Q1.  Let (a,b,c) be distinct non-zero real numbers satisfying a + 2/b = b + 2/c = c + 2/a. Determine the value of |a^2.b + b^2.c + c^2.a|. S1. The fact that a,b,c are given as distinct => problem wants us to cancel out factors like a-b,b-c,c-a. a-b = 2(c-b)/bc similarly create other equations and multiply all to get (abc)^2 = 8 Then again simplify the given equations to get a^2b + ... = 3abc => answer = 6.sqrt(2) Q2. Let (S) be the set of all three-digit positive integers (abc) (where (a,b,c) are the digits) such that the two-digit number (ab) is divisible by (c), and the two-digit number (bc) is divisible by (a). If we exclude the trivial cases (a=b=c), what is the largest three-digit integer in set (S)? S2. a | bc and c | ab To make it largest, try a = 9 9 | bc c | 9b Now let's try values of bc bc = 99 trivial bc = 90 c can't be 0 as c | ab fails bc = 81 works since 1 | 98 So answer = 981 Q3. Find infinitely many triples ((a,b,c)) of positive integers such that (a, b, c...
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practice problems pending from Q3

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Q1. Use mathematical induction to derive the following formula for all n>=1: 1.(1!) + 2.(2!) + 3.(3!) .... n.(n!) = (n+1)! - 1 S1. Pr. th. (n+1)(n+1)! + (n+1)! - 1 = (n+2)! - 1 => (n+1)!(n+2) - 1 = (n+2)! - 1 => (n+2)! = (n+2)! H.P. Q2. + 2 ( 2 ! ) + 3 ( 3 ! ) + ⋯ + n ( n ! ) = ( n + 1 ) S2. (a) Assuming it holds for 'n', i.e. 1/1^2 ... + 1/n^2 <= 2 - 1/n 1/1^2 ... + 1/(n+1)^2 <= 2 - 1/(n+1) If we can show that: 2 - 1/n + 1/(n+1)^2 <= 2 - 1/(n+1) we are done. If: 2 - 1/n + 1/(n+1)^2 <= 2 - 1/(n+1) Then: - 1/n + 1/(n+1)^2 <= - 1/(n+1) then: 1/(n+1)^2 <= 1/n - 1/(n+1) => 1/(n+1)^2 <= 1/n(n+1) which is true. H.P. (b) Pr. th. (n+1)/2^(n+1) + 2 - (n+2)/2^n = 2 - (n+3)/2^(n+1) => (n+1)/2^(n+1) - (n+2)/2^n = - (n+3)/2^(n+1) => (n+1)/2 - (n+2) = -(n+3)/2 => n + 2 = (2n+4)/2 H.P. Q3. S3. Using the second principle of induction, assume that the given relation holds for n,n-1,n-2, then: a_(n-1) = 5.2^(n-1) + 1 a_(n-2) = 5.2^(n-2) + 1 So a_n = ...
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