Prove that 8 times any triangle number is always 1 less than a square number
A triangle number is a number such that it is the sum of n consecutive integers, starting from 0. Eg 1, 1+2, 1+2+3... are the first 3 triangle numbers. The formula for the nth triangle number is well-known at A-level and is (1/2)(n)(n+1); the formula for the sum of the first n integers. To answer the question we must show that 8N+1, where N is any triangle number, is a perfect square.
8N+1 = 8(1/2)(n)(n+1)+1 = 4n(n+1)+1=4n^2 + 4n + 1This is a perfect square, as we can rewrite it as:(2n+1)^2with it's root of 2n+1 being an integer. Therefore we have shown that 8N+1 is a perfect square, hence the result has been proved.
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