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This method can be applied to problem #6 at IMO 1988: Let a and b be positive integers such that ab + 1 divides a 2 + b 2. Prove that a 2 + b 2 / ab + 1 is a perfect square. Let a 2 + b 2 / ab + 1 = q and fix the value of q. If q = 1, q is a perfect square as desired.
A perfect square is an element of algebraic structure that is equal to the square of another element. ... Perfect square trinomials, a method of factoring polynomials
The polynomial x 2 + cx + d, where a + b = c and ab = d, can be factorized into (x + a)(x + b).. In mathematics, factorization (or factorisation, see English spelling differences) or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.
Pell's equation for n = 2 and six of its integer solutions. Pell's equation, also called the Pell–Fermat equation, is any Diophantine equation of the form =, where n is a given positive nonsquare integer, and integer solutions are sought for x and y.
Brocard's problem is a problem in mathematics that seeks integer values of such that ! + is a perfect square, where ! is the factorial. Only three values of n {\displaystyle n} are known — 4, 5, 7 — and it is not known whether there are any more.
Layers of Pascal's pyramid derived from coefficients in an upside-down ternary plot of the terms in the expansions of the powers of a trinomial – the number of terms is clearly a triangular number. In mathematics, a trinomial expansion is the expansion of a power of a sum of three terms into monomials. The expansion is given by
This is comparable with the 24th square pyramid having a total of 70 2 cannonballs. [5] Similarly, a pentagonal-pyramid version of the cannonball problem to produce a perfect square, would have N = 8, yielding a total of (14 × 14 = ) 196 cannonballs. [6] The only numbers that are simultaneously triangular and square pyramidal are 1, 55, 91 ...
The effect has been to fold up the u 4 term into a perfect square: (u 2 + a) 2. The second term, au 2 did not disappear, but its sign has changed and it has been moved to the right side. The next step is to insert a variable y into the perfect square on the left side of equation , and a corresponding 2y into the coefficient of u 2 in the