Suppose that \(f(x)\) is a quadratic real polynomial. If for any positive integer \(n\), \(f(n)\) is a square of an integer, prove that \(f(x)\) is a square of a one degree polynomial with integer coefficients.

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TopNewestConsider the quadratic function in vertex form: f(x) = a(x-h)^2 + k. We express this as a perfect square: a(x - h)^2 + k = m^2.

k can be expressed as: m^2 - a(x - h)^2 = k.

Now, this is one Pell-type equation.

Case 1: If a is not a perfect square and k is not equal to 0. Consider the behavior of a Pell-type equation where the solutions are exponentially increasing which implies that the domain is not all positive integers for its range to become a perfect square.

Case 2: If a is not a perfect square and has norm 0. m cannot be integral due to the irrational number a.

(Let a = n^2 for cases 3 and 4.) Case 3: If a is a perfect square and k has norm not equal to 0. k = (m - n(x-h))(m + n(x-h)) In order for m, x, h, and n be all integral, the main factors must be equated to the factors of k but its factors are finite so it cannot be guaranteed that its domain is all positive integers.

Case 4: If a is a perfect square and k has norm equal to 0. m = n(x - h) m = -n(x - h)

Exhausting all cases, case 4 satisfies the conditions. Hence, f(x) is a linear degree polynomial with integer coefficients. – John Ashley Capellan · 2 years, 11 months ago

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– Bogdan Simeonov · 2 years, 3 months ago

The problem says that it's a real quadratic functions, so the coefficients could be any real number, but you use Pell's equation, which depends on them being integers.Log in to reply

– John Ashley Capellan · 2 years, 11 months ago

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WELL, LET POLYNOMIAL BE ax^2+bx+c THEN c IS A QUADRATIC RESIDUE FOR EVERY NATURAL VALUE OF x HENCE c IS A PERFECT SQUARE. LET c BE m^2 THEN f(x) =ax^2+bx+m^2 Now IT CAN BE SEEN THAT b IS A MULTIPLE OF EVERY PRIME FACTOR OF m

I KNOW THIS IS NOT A COMPLETE SOLUTION BUT I BELIEVE IT WILL HELP – Subrata Saha · 2 years, 12 months ago

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