Square Of A One Degree Polynomial With Integer Coefficients
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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Consider 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.
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.
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
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Consider 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.
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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.
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