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Let ( n ) be an even positive integer. Let ( p ) be a monic, real polynomial of degree ( 2n ); that is to say,

[ p(x) = x^{2n} + a_{2n-1}x^{2n-1} + \dots + a_1x + a_0 ]

for some real coefficients ( a_0, \dots, a_{2n-1} ).

Suppose that ( p\left( \frac{1}{k} \right) = k^2 ) for all integers ( k ) such that ( 1 \leq |k| \leq n ).

Find all other real numbers ( x ) for which ( p\left( \frac{1}{x} \right) = x^2 ).

Assistant

Chain of Thought

We are told that for all integer values of ( k ) satisfying ( 1 \leq |k| \leq n ), [ p\left(\frac{1}{k}\right) = k^2 ] Our problem is: Given that ( p ) is monic of degree ( 2n ), and that ( p\left(\frac{1}{k}\right) = k^2 ) for ( k = \pm1, \pm2, \dots, \pm n ), find all other real numbers ( x ) such that [ p\left(\frac{1}{x}\right) = x^2. ]

In other words, we are to find all real numbers ( x ) such that ( p\left(\frac{1}{x}\right) = x^2 ).

Firstly, note that:

• The monic polynomial ( p(z) ) is of degree ( 2n ).

• We have ( 2n ) data points: for ( k = \pm1 \text{ to } \pm n, p\left(\frac{1}{k}\right) = k^2 ).

But there are ( 2n ) distinct values of ( \frac{1}{k} ) (since ( k ) ranges over integers ( \pm1 \text{ to } \pm n ), and they are distinct).

Thus, we have ( 2n ) points ( (x, y) ) of the form ( \left( \frac{1}{k}, k^2 \right) ).

Thus, ( p ) is a polynomial of degree ( 2n ) that satisfies ( p\left(\frac{1}{k}\right) = k^2 ) for ( 2n ) values of ( \frac{1}{k} ).

Wait, but ( p\left(\frac{1}{x}\right) - x^2 = 0 ).

Our equation is ( p\left(\frac{1}{x}\right) - x^2 = 0 ).

Let me define ( q(x) = p\left(\frac{1}{x}\right) - x^2 ).

Our task is to find all real numbers ( x ) such that ( q(x) = 0 ).

Given the data from the problem, let’s consider that ( q(k) = p\left(\frac{1}{k}\right) - k^2 = 0 ) for ( k = \pm1, \pm2, \dots, \pm n ).

Thus, ( q(k) = 0 ) for ( k = \pm1, \dots, \pm n ).

Moreover, since ( p ) is a polynomial of degree ( 2n ), then ( p\left(\frac{1}{x}\right) ) is a rational function of ( x ) of degree ( 2n ) when considered as a function of ( x ).

But we can consider ( q(x) = p\left(\frac{1}{x}\right) - x^2 ).

Alternatively, let’s consider ( s(x) = x^{2n} p\left(\frac{1}{x}\right) - x^{2n+2} ).

Let me explain: We can consider that ( p\left(\frac{1}{x}\right) = x^{-2n} + a_{2n-1}x^{-2n+1} + \dots + a_0 ).

Then multiplying by ( x^{2n} ) gives: [ x^{2n} p\left(\frac{1}{x}\right) = 1 + a_{2n-1}x + a_{2n-2}x^2 + \dots + a_0 x^{2n}. ] Similarly, ( x^{2n+2} = x^{2n+2} ).

Thus, define the polynomial ( s(x) = x^{2n} p\left(\frac{1}{x}\right) - x^{2n+2} ).

But wait, perhaps it’s better to rearrange.

Let me consider that ( s(x) = x^{2n}\left[ p\left(\frac{1}{x}\right) - x^2 \right] = x^{2n} q(x) ).

So ( s(x) = x^{2n} q(x) ).

Since ( q(k) = 0 ) for ( k = \pm1, \dots, \pm n ), we have ( q(x) = 0 ).

Thus, ( q(x) = k^{2n} q(x) = 0 ).

Thus, ( q(k) = 0 ) at ( x = \pm1, \pm2, \dots, \pm n ).

Thus, ( s(x) ) is a polynomial (not a rational function), since ( x^{2n} q(x) ) is a polynomial.

Moreover, ( s(x) = x^{2n} \left[ p\left(\frac{1}{x}\right) - x^2 \right] = x^{2n} p\left(\frac{1}{x}\right) - x^{2n+2} ).

But ( x^{2n} p\left(\frac{1}{x}\right) = p^*(x) ).

Wait, but ( x^{2n} p\left(\frac{1}{x}\right) ) is the "reciprocal polynomial" of ( p(x) ), evaluated at ( x ).

But since ( p(x) ) is monic of degree ( 2n ), its reciprocal polynomial is monic of degree ( 2n ) as well.

Thus, set ( t(x) = x^{2n} p\left(\frac{1}{x}\right) ).

Let me write ( t(x) = x^{2n} p\left(\frac{1}{x}\right) = P^*(x) ).

( P^*(x) ) is the reverse polynomial of ( p(x) ).

( p(x) = x^{2n} + a_{2n-1}x^{2n-1} + \dots + a_0 ).

Then ( t(x) = x^{2n} p\left(\frac{1}{x}\right) = x^{2n} \left[ \frac{1}{x^{2n}} + a_{2n-1} \frac{1}{x^{2n-1}} + \dots + a_0 \right] = 1 + a_{2n-1} x + a_{2n-2} x^2 + \dots + a_0 x^{2n} ).

So ( t(x) = p^*(x) ), the reciprocal polynomial of ( p ).

Thus, ( s(x) = t(x) - x^{2n+2} = 1 + a_{2n-1}x + a_{2n-2}x^2 + \dots + a_0 x^{2n} - x^{2n+2} ).

But ( x^{2n+2} = x^{2n+2} ).

Therefore, ( s(x) = p^*(x) - x^{2n+2} ).

We can now write ( s(x) = p^*(x) - x^{2n+2} ).

Alternatively, ( p^*(x) - x^{2n+2} = s(x) ).

But ( t(x) = p^*(x) ), and ( s(x) = t(x) - x^{2n+2} ).

But our key is that ( s(x) = x^{2n} q(x) ).

Recall that ( q(k) = 0 ) for ( k = \pm1, \pm2, \dots, \pm n ).

Therefore, ( s(k) = 0 ) at ( x = \pm1, \dots, \pm n ).

Therefore, ( s(x) ) is a polynomial that vanishes at ( x = \pm1, \pm2, \dots, \pm n ).

( s(x) ) has roots at ( x = \pm1, \dots, \pm n ).

Thus, ( s(x) ) must be divisible by the monic polynomial whose roots are at ( \pm1, \pm2, \dots, \pm n ).

Thus, ( s(x) = m(x)h(x) ).

Where ( m(x) = (x - (-n))(x - (-n + 1)) \dots (x - (2 - 1))(x - (1))(x - n) ).

Wait, that is, ( m(x) = \prod_{k=1}^n(x - k)(x + k) = \prod_{k=1}^n(x^2 - k^2) ).

Therefore, ( m(x) = \prod_{k=1}^n(x^2 - k^2) ).

Thus, ( m(x) ) is a monic polynomial of degree ( 2n ).

Similarly, since ( s(x) ) is of degree...

Let me compute the degree of ( s(x) ):

Since ( s(x) = x^{2n} p\left(\frac{1}{x}\right) - x^{2n+2} ),

Compute ( t(x) = x^{2n} p\left(\frac{1}{x}\right) ).

We have ( p(z) = x^{2n} + a_{2n-1}x^{2n-1} + \dots + a_0 ).

Then ( t(x) = x^{2n} p\left(\frac{1}{x}\right) = x^{2n} \left[ \frac{1}{x^{2n}} + a_{2n-1}\frac{1}{x^{2n-1}} + \dots + a_0 \right] = 1 + a_{2n-1}x + a_{2n-2}x^2 + \dots + a_0 x^{2n} ).

So ( t(x) ) is a polynomial of degree ( 2n ).

What about ( x^{2n+2} )?

So ( s(x) = t(x) - x^{2n+2} ).

Therefore, ( s(x) ) is a polynomial of degree ( \max(\text{degree of } t(x), \text{degree of } x^{2n+2}) = \text{degree of } x^{2n+2} ).

So ( s(x) ) is of degree ( 2n + 2 ).

But ( t(x) ) has degree ( 2n ).

So ( s(x) = [\text{degree } 2n \text{ polynomial}] - [\text{degree } (2n + 2) \text{ monomial}] = -x^{2n+2} + [\text{degree } \leq 2n \text{ polynomial}] ).

So the leading term of ( s(x) ) is ( -x^{2n+2} ).

Thus, ( s(x) ) is a polynomial of degree ( 2n + 2 ) with leading coefficient ( -1 ).

So ( s(x) = -x^{2n+2} + \text{lower degree terms} ).

Similarly, ( m(x) = \prod_{k=1}^n \left( x^2 - k^2 \right) = \text{monic polynomial of degree } 2n ).

Therefore, if we factor ( s(x) = m(x)h(x) ).

Since ( s(x) ) is degree ( 2n + 2 ) and ( m(x) ) is degree ( 2n ), ( h(x) ) must be degree 2.

Therefore, ( h(x) ) is a quadratic polynomial.

Our plan is to write ( s(x) = m(x)h(x) ).

Given that ( s(x) ) is degree ( 2n + 2 ) with leading coefficient ( -1 ).

Similarly, ( m(x) ) is a degree ( 2n ) monic polynomial.

Therefore, ( h(x) ) must be a degree 2 polynomial with leading coefficient ( -1 ).

Therefore, ( h(x) ) must be of the form ( h(x) = -x^2 + bx + c ).

Alternatively, since ( s(x) = m(x)h(x) ),

We can write ( s(x) = (-1)x^{2n+2} + [\text{lower degree terms}] ).

Also, ( m(x) = x^{2n} + [\text{lower degree terms}] ) (since it is monic of degree ( 2n )).

Then ( s(x) = m(x)h(x) = \left[ x^{2n} + \dots \right](-x^2 + bx + c) = -x^{2n+2} + [\text{lower degree terms}] ).

Therefore, the leading term of ( s(x) ) is ( -x^{2n+2} ), which matches.

Therefore, our assertion that ( h(x) ) is degree 2 with leading coefficient ( -1 ) is consistent.

Therefore, ( h(x) = -x^2 + bx + c ).

Our task now is to find ( b ) and ( c ).

So we have:

[ s(x) = m(x)h(x) = \left[ \prod_{k=1}^n (x^2 - k^2) \right] (-x^2 + bx + c) ]

and

[ s(x) = x^{2n} p\left(\frac{1}{x}\right) - x^{2n+2}. ]

Equate the expressions for ( s(x) ).

Alternatively, let’s consider that ( t(x) = x^{2n} p\left(\frac{1}{x}\right) = p^*(x) ).

Then ( s(x) = t(x) - x^{2n+2} = p^*(x) - x^{2n+2} ).

Therefore, ( s(x) = m(x)h(x) ).

Thus, ( p^*(x) - x^{2n+2} = m(x)h(x) ).

But ( p^*(x) ) is degree ( 2n ) and ( x^{2n+2} ) is degree ( 2n + 2 ), so their difference is degree ( 2n + 2 ).

Therefore, the left-hand side ( p^*(x) - x^{2n+2} = [\text{degree} \leq 2n] - [\text{degree } 2n + 2] ).

Wait, but the ( x^{2n+2} ) term is of higher degree than ( p^*(x) ), so the leading term of ( s(x) ) is ( -x^{2n+2} ).

Similarly, the right-hand side ( m(x)h(x) = [\text{degree } 2n] \times [\text{degree } 2] = [\text{degree } 2n + 2] ).

Now, let’s write ( s(x) = -x^{2n+2} + \text{lower degree terms} ).

Similarly, ( m(x)h(x) = x^{2n}(-x^2 + bx + c) + \text{lower degree terms} ).

Compute ( m(x)h(x) = \left( x^{2n} + \text{lower degree terms} \right) (-x^2 + bx + c) ).

The product of ( x^{2n} ) and ( -x^2 ) gives ( -x^{2n+2} ).

Similarly, ( x^{2n} \cdot bx = bx^{2n+1} ).

And ( x^{2n} \cdot c = cx^{2n} ).

Thus, we find:

[ s(x) = m(x)h(x) = -x^{2n+2} + bx^{2n+1} + cx^{2n} + \text{[lower degree terms]}. ]

Similarly, ( s(x) = -x^{2n+2} + [\text{lower degree terms}] ).

Thus, to make the higher-degree terms match, ( s(x) = -x^{2n+2} + bx^{2n+1} + cx^{2n} + [\text{lower degree terms}] ).

Wait, but from the definition of ( s(x) ), we have:

( s(x) = [t(x)] - x^{2n+2} = [\text{leading terms of } t(x)] - x^{2n+2} ).

But ( t(x) = p^*(x) = 1 + a_{2n-1}x + a_{2n-2}x^2 + \dots + a_0x^{2n} ).

Thus, ( t(x) = ) polynomial of degree ( \leq 2n ).

Therefore, ( s(x) = [\text{degree } \leq 2n] - x^{2n+2} = -x^{2n+2} + [\text{degree } \leq 2n] ).

Therefore, the highest degree term of ( s(x) ) is ( -x^{2n+2} ), and the rest is lower degree terms.

Therefore, ( s(x) = -x^{2n+2} + \text{lower degree terms} ).

So matching terms:

From ( m(x)h(x) ):

• Leading term of ( s(x) = -x^{2n+2} )
• Second term: ( bx^{2n+1} )
• Third term: ( cx^{2n} )

But from ( s(x) = p^*(x) - x^{2n+2} ), and ( p(x) ) has degree ( \leq 2n ).

Therefore, there is no ( x^{2n+1} ) term in ( s(x) ) unless ( t(x) ) contributes, but ( t(x) ) is of degree ( \leq 2n ).

Given that ( s(x) = -x^{2n+2} + [\text{degree } \leq 2n] ),

Therefore, ( s(x) = -x^{2n+2} + bx^{2n+1} + a_{2n-1}x^{2n-1} + \dots + a_0 ).

Similarly, from the ( m(x)h(x) ) computation:

[ s(x) = m(x)h(x) = \left[ x^{2n} + \dots \right](-x^2 + bx + c) = -x^{2n+2} + bx^{2n+1} + cx^{2n} + [\text{lower degree terms}] ]

So from the above, the coefficient of ( x^{2n+1} ) in ( s(x) ) is ( b ).

But according to ( s(x) = -x^{2n+2} + [\text{degree } \leq 2n] ), the ( x^{2n+1} ) term is absent (since ( p^*(x) ) has degree ( \leq 2n )).

Therefore, ( s(x) ) has no ( x^{2n+1} ) term.

Thus, the coefficient of ( x^{2n+1} ) in ( s(x) ) is zero.

Therefore, ( b = 0 ).

Similarly, let's now compare the coefficient of ( x^{2n} ) in ( s(x) ), which is ( cx^{2n} ).

Similarly, from ( s(x) = -x^{2n+2} + bx^{2n+1} + [\text{lower degree terms}] ),

So the coefficient of ( x^{2n} ) in ( s(x) ) is ( cx^{2n} = c ).

But ( s(x) = p^*(x) - x^{2n+2} ).

So ( s(x) = [p^*(x)] - x^{2n+2} ).

But ( p^*(x) ) has degree ( \leq 2n ).

Therefore, the ( 2n ) term in ( s(x) ) is given by the ( 2n ) term in ( p^*(x) ), which is ( a_0x^{2n} ).

So ( s(x) = (-x^{2n+2}) + a_0x^{2n} + [\text{lower degree terms}] ).

Therefore, comparing coefficients, the coefficient of ( x^{2n} ) in ( s(x) ) is ( a_0 ).

Thus, ( c = a_0 ).

So ( c = a_0 ).

So we have ( h(x) = -x^2 + c ).

But we already have ( b = 0 ).

Thus, ( h(x) = -x^2 + c ).

We can now write: ( s(x) = m(x)[-x^2 + c] = \left[ \prod_{k=1}^n (x^2 - k^2) \right] (-x^2 + c) ).

Similarly, ( s(x) = -x^{2n+2} + a_0x^{2n} + [\text{lower degree terms}] ).

We may be able to compute the expression ( m(x)[-x^2 + c] ) and match the coefficient of ( x^{2n} ).

Let me first consider ( m(x) = x^{2n} + [\text{lower degree terms}] ).

Similarly, ( m(x) \cdot (-x^2 + c) = -x^{2n+2} + cx^{2n} + [\text{lower degree terms}] ).

Therefore, ( s(x) = -x^{2n+2} + cx^{2n} + [\text{lower degree terms}] ).

But ( s(x) = -x^{2n+2} + a_0x^{2n} + [\text{lower degree terms}] ).

Therefore, ( c = a_0 ).

Therefore, ( c = a_0 ).

Thus, ( c = a_0 ).

So we’ve confirmed that ( c = a_0 ).

But what about the next term? Let’s consider matching the next coefficient.

Compute ( m(x) \cdot (-x^2 + c) = -x^{2n+2} + cx^{2n} + [\text{lower degree terms}] ).

The term of degree ( 2n - 2 ) in ( s(x) ) is due to the products:

• The ( x^{2n - 2} ) term from ( m(x) ) multiplied by ( -x^2 ).
• The ( x^{2n - 2} ) term from ( m(x) ) multiplied by ( c ).

Let me write ( m(x) = x^{2n} + m_{2n-2}x^{2n-2} + \text{lower terms} ).

Similarly, ( m(x) \cdot (-x^2) = (-x^{2n+2} + m_{2n-2}x^{2n} + \text{lower terms}) ).

Wait, actually, this may not help us directly without specific terms.

An alternative is to consider the known polynomials.

So ( m(x) = \prod_{k=1}^n (x^2 - k^2) ).

Similarly, ( s(x) = m(x)[-x^2 + c] ).

Now, consider that ( s(x) = [-x^{2n+2}] + cx^{2n} + [\text{lower degree terms}] ).

Similarly, expand ( m(x)[-x^2] ) and ( m(x) \cdot c ) separately.

Let me compute ( m(x) \cdot (-x^2) ).

Since ( m(x) ) is a monic degree ( 2n ) polynomial, its highest degree term is ( x^{2n} ).

Therefore, ( m(x) \cdot (-x^2) = -x^{2n+2} + \text{lower degree terms} ).

Similarly, ( m(x) \cdot c = cx^{2n} + \text{lower degree terms} ).

Therefore, ( s(x) = [-x^{2n+2}] + cx^{2n} + [\text{lower degree terms}] ).

Similarly, from the definition of ( s(x) ), ( s(x) = [-x^{2n+2}] + a_0x^{2n} + a_{2n-1}x^{2n-1} + \dots ).

But from ( a_0 = p^*(x) - x^{2n+2} ),

Since ( p^*(x) = 1 + a_{2n-1}x + a_{2n-2}x^2 + \dots + a_0x^{2n} ),

Thus, ( s(x) = [1 + a_{2n-1}x + a_{2n-2}x^2 + \dots + a_0x^{2n}] - x^{2n+2} ).

But this seems to contradict our prior assertion that ( s(x) = -x^{2n+2} + a_0x^{2n} + \dots ).

Wait, perhaps we need to be careful here.

Let me try to rewrite ( s(x) ).

Given that ( t(x) = p^*(x) = x^{2n}p\left(\frac{1}{x}\right) = 1 + a_{2n-1}x + \dots + a_0x^{2n} ),

Therefore, ( s(x) = t(x) - x^{2n+2} = [1 + a_{2n-1}x + \dots + a_0x^{2n}] - x^{2n+2} ).

So ( s(x) = -x^{2n+2} + a_0x^{2n} + [\text{lower degree terms}] ).

So the constant term of ( s(x) ) is ( s(0) = p^(0) - 0 = p^(0) = 1 ).

Similarly, ( m(x) \cdot h(x) ), when evaluated at ( x = 0 ), becomes ( m(0) \cdot h(0) = \left[ \prod_{k=1}^n (1 - k^2) \right] [-0 + c] = \left[ (-1)^n k_1^2 k_2^2 \dots k_n^2 \right] c ).

But since ( k = 1 \text{ to } n ), the product ( k^2 ) is

Wait, sorry, actually, ( \prod_{k=1}^n k^2 = (n!)^2 )

So ( m(0) = (-1)^n(n!)^2 ).

Therefore, ( m(0)h(0) = (-1)^n(n!)^2(0 + c) = (-1)^n(n!)^2c ).

Similarly, ( s(0) = 1 ).

So we have:

[ s(0) = m(0)h(0). ]

[ 1 = (-1)^n(n!)^2c. ]

Therefore,

[ c = \frac{1}{(-1)^n(n!)^2} = (-1)^n \frac{1}{(n!)^2}. ]

Wait, but ( c = a_0 ).

But we previously had ( c = a_0 ).

Therefore, ( a_0 = (-1)^n \frac{1}{(n!)^2} ).

But since ( c = a_0 ),

Thus, we have determined ( c ).

Similarly, perhaps we can find more about ( p^*(x) ).

But perhaps we can do better.

Alternatively, since ( p^*(x) = -x^{2n+2} = s(x) = m(x)h(x) ).

Given that ( m(x) = \prod_{k=1}^n (x^2 - k^2) = x^{2n} - (\text{sum of squares})x^{2n-2} + \dots + (-1)^n(n!)^2 ).

Similarly, ( t(x) = p^*(x) ).

Given ( p^*(x) = x^{2n} p\left( \frac{1}{x} \right) = x^{2n} \left(1 + a_{2n-1} \frac{1}{x} + a_{2n-2} \frac{1}{x^2} + \dots + a_0 \right) ).

Therefore, ( p^*(x) = 1 + a_{2n-1}x + a_{2n-2}x^2 + \dots + a_0 x^{2n} ).

Thus, ( p^*(x) ) is degree ( 2n ).

Similarly, ( s(x) = p^*(x) - x^{2n+2} ).

Now, what if we consider that ( p^*(x) = x^{2n+2} - m(x)h(x) ).

Wait, but that seems messy.

Alternatively, since ( s(x) = m(x)h(x) = [x^{2n} + \dots + (-1)^n(n!)^2](-x^2 + c) ).

To find ( p^*(x) ), we can write

[ p^*(x) = x^{2n+2} - [m(x)x^2 + cm(x)] + x^{2n+2}. ]

Thus, ( p^*(x) = [m(x)x^2 + cm(x)] + x^{2n+2} ).

Then, ( p^*(x) = [m(x)x^2 + cm(x)] + x^{2n+2} ).

So ( p^*(x) = [m(x)(-x^2 + c)] + x^{2n+2} ).

But ( x^{2n+2} = x^2x^{2n} ).

But ( m(x) ) is degree ( 2n ), and we can write ( x^{2n+2} = x^2a_0x^{2n} ).

So ( p^*(x) = m(x)x^2 + cm(x) + x^{2n+2} ).

But ( m(x)x^2 = a_2m(x) = x^{2n} + \dots ).

So let’s compute ( p^*(x) = -x^2m(x) + cm(x) + x^{2n+2} ).

But ( x^{2n+2} = x^2a_2 ), so:

[ p^*(x) = -x^2m(x) + cm(x) + x^{2n+2}. ]

Now ( m(x) ) is monic degree 2n: ( m(x) = x^{2n} + \dots + (-1)^n(n!)^2 ).

So ( 2x^2m(x) = x^{2n} + \text{lower degree terms} ).

Therefore, ( -x^2m(x) = -x^{2n+2} + \text{lower degree terms} ).

Similarly, ( a_2x^2 = x^{2n+2} ).

So ( p^*(x) = -x^2m(x) + cm(x) + x^{2n+2} = [-x^2m(x) + x^{2n}] + cm(x) ).

But ( x^2m(x) = x^{2n+2} + a_2m(x) ) (since ( m(0) ) is the leading coefficient of ( x^{2n} ), which is 1).

Therefore, ( -x^2m(x) + x^{2n+2} = -x^2[m(x) - x^{2n}] ).

But ( m(x) = x^{2n} + \text{lower degree terms} ).

Therefore, ( -x^2m(x) + x^{2n+2} = -x^2[m(x) - x^{2n}] ).

Thus, ( p^*(x) = -x^2[m(x) - x^{2n}] + cm(x) ).

Therefore, ( p^*(x) = -x^2[m(x) - x^{2n}] + cm(x) ). But ( m(x) - x^{2n} = \text{lower degree terms} ).

Similarly, ( p^*(x) = -x^2[\text{lower degree terms}] + cm(x) ).

So the term of degree ( 2n + 1 ) in ( p^*(x) ) comes from ( -x^2[m_{2n-1}x^{2n-1}] ) plus ( cm(x) ).

But ( p^*(x) ) has degree ( \leq 2n ).

Therefore, the degree of ( p^*(x) ) is ( \leq 2n ).

Similarly, the highest degree term in ( p^*(x) ) is ( x^{2n} ).

But from our prior expression, ( p^*(x) = -x^2[m(x) - x^{2n}] + cm(x) ).

But after simplification, the leading term is ( cx^{2n} ).

Wait, I think this is getting too messy.

Perhaps I should consider the concrete value of ( m(x) ) at ( x = 0 ).

We already considered

[ s(0) = 1. ]

Similarly, ( m(0) = (-1)^n(n!)^2 ).

Therefore, we established that ( 1 = s(0) = m(0)h(0) ).

But ( h(0) = 0 + c = c ).

So

[ c = \frac{1}{m(0)} = \frac{1}{(-1)^n(n!)^2} = (-1)^n \frac{1}{(n!)^2}. ]

So ( c = \frac{(-1)^n}{(n!)^2} ).

Similarly, we had ( c = a_0 ).

Therefore, ( a_0 = c = \frac{(-1)^n}{(n!)^2} ).

Therefore, the constant term in ( p^(x) ) is ( p^(0) = 1 ).

So we have ( p^*(0) = 1 ).

Similarly, ( s(0) = p^*(0) - 0 = 1 ).

So consistent.

But perhaps this is as far as we can go without specific computations.

Alternatively, the problem is to find all real numbers ( x ) satisfying

[ p\left( \frac{1}{x} \right) = x^2. ]

Given that ( p\left( \frac{1}{k} \right) = k^2 ) for ( k = \pm 1, \pm 2, \dots, \pm n ),

We might conjecture that the only real solutions are ( x = \pm (n+1), \pm (n+2), \dots ).

But since ( p\left( \frac{1}{x} \right) - x^2 = 0 ), and ( p\left( 1/k \right) ) is given, perhaps all solutions ( x ) satisfy ( p\left( 1/x \right) = x^2 ).

But we can express ( q(x) = p\left( 1/x \right) - x^2 ).

But ( q(x) ) has roots at ( x = \pm 1, \pm 2, \dots, \pm n ).

Similarly, the solutions we are looking for are zeros of ( q(x) ).

Thus, ( q(x) = \frac{s(x)}{x^2} ).

But ( s(x) = x^{2n}q(x) ).

Similarly, ( s(x) = m(x)h(x) ), where ( h(x) = -x^2 + c ).

So the zeros of ( q(x) ) are zeros of ( s(x) ) of degree ( 2n + 2 ) minus zeros at ( x = 0 ) multiplicity ( 2n ) (since ( s(x) = x^{2n}q(x) )).

Therefore, zeros of ( q(x) ) are zeros of ( s(x) ), excluding ( x = 0 ).

Thus, zeros of ( q(x) ) are ( x = \pm 1, \dots, \pm n ), and zeros of ( h(x) ).

But ( h(x) = -x^2 + c = 0 ).

So the other zeros are solutions of ( -x^2 + c = 0 ).

So solving ( -x^2 + c = 0 ).

That is, ( x^2 = c ).

Thus, ( x = \pm \sqrt{c} ).

But ( c = \frac{(-1)^n}{(n!)^2} ).

So

[ c = \frac{(-1)^n}{(n!)^2}. ]

Therefore, ( c > 0 ) when ( n ) is even, since ( (-1)^n = 1 ).

Therefore, when ( n ) is even, ( c = \frac{1}{(n!)^2} > 0 ).

Thus, ( x^2 = c ) has real solutions ( x = \pm \sqrt{c} ).

Therefore, the other real solutions of ( p\left( \frac{1}{x} \right) = x^2 ) are ( x = \pm \sqrt{c} ).

Similarly, when ( n ) is even, ( c > 0 ).

Thus, the other real numbers ( x ) satisfying ( p\left( \frac{1}{x} \right) = x^2 ) are ( x = \pm \frac{1}{n+1} ).

Therefore, the other real numbers are ( x = \pm \frac{1}{n+1} ).

Wait, wait, ( x^2 = c ), so ( x = \pm \sqrt{\frac{1}{(n!)^2}} = \pm \frac{1}{n!} ).

Therefore, the other real numbers are ( x = \pm \frac{1}{n!} ).

So in conclusion, the numbers ( x ) satisfying ( p\left( \frac{1}{x} \right) = x^2 ) are ( x = \pm k ) for ( k = 1 \text{ to } n ), and ( x = \pm \frac{1}{n!} ).

Thus, the other real numbers are ( x = \pm \frac{1}{n!} ).

So the answer is ( x = \pm \frac{1}{n!} ) (along with ( x = \pm k \text{ for } k = 1 \text{ to } n )).

Wait, but the problem says “Suppose that ( p\left( \frac{1}{k} \right) = k^2 ) for all integers ( k ) with ( 1 \leq |k| \leq n ). Find all other real numbers for which ( p\left( \frac{1}{x} \right) = x^2 )."

Thus, apart from ( x = \pm k ) (which have already been given), the other real numbers are ( x = \pm \frac{1}{n!} ).

Wait, but when ( n ) is even, ( c = \frac{(-1)^n}{(n!)^2} = \frac{1}{(n!)^2} ).

Thus, ( c = \frac{1}{(n!)^2} ).

Thus, ( x^2 = c \Rightarrow x = \pm \frac{1}{n!} ).

Thus, the other real numbers are ( x = \pm \frac{1}{n!} ).

Answer: The other real numbers ( x ) satisfying ( p\left( \frac{1}{x} \right) = x^2 ) are ( x = \pm \frac{1}{n!} ).

Answer

To solve this problem, we need to find all real numbers ( x ) such that

[ p\left( \frac{1}{x} \right) = x^2, ]

where ( p ) is a monic real polynomial of degree ( 2n ), and it is given that

[ p\left( \frac{1}{k} \right) = k^2 ]

for all integers ( k ) with ( 1 \leq |k| \leq n ).

Understanding the Given Information:

1. Given Data Points: We have ( 2n ) data points from ( k = \pm 1, \pm 2, \dots, \pm n ), which provide the values ( p\left( \frac{1}{k} \right) = k^2 ).

2. Polynomial Function: ( p(x) ) is monic of degree ( 2n ), which means its highest-degree coefficient is 1.

Defining a New Polynomial:

Let's define a new function

[ q(x) = p\left( \frac{1}{x} \right) - x^2. ]

Our goal is to find all real solutions to ( q(x) = 0 ).

Properties of ( q(x) ):

• Since ( p ) is a polynomial of degree ( 2n ), ( p\left( \frac{1}{x} \right) ) can be expressed in powers of ( x ).
• The function ( q(x) ) has zeroes at ( x = \pm 1, \pm 2, \dots, \pm n ) because ( q(k) = p\left( \frac{1}{k} \right) - k^2 = 0 ).

Constructing ( s(x) ):

Consider the polynomial

[ s(x) = x^{2n} q(x) = x^{2n} \left( p\left( \frac{1}{x} \right) - x^2 \right). ]

Multiplying through by ( x^{2n} ) eliminates denominators, making ( s(x) ) a polynomial.

Factoring ( s(x) ):

We know that ( s(x) ) has roots at ( x = \pm 1, \pm 2, \dots, \pm n ). Therefore, ( s(x) ) can be expressed as:

( s(x) = m(x)h(x) ),

where ( m(x) = \prod_{k=1}^n (x^2 - k^2) ) is a monic polynomial of degree ( 2n ) whose roots are ( \pm 1, \pm 2, \dots, \pm n ), and ( h(x) ) is a quadratic polynomial.

Determining ( h(x) ):

We consider ( h(x) ) to be a quadratic polynomial:

[ h(x) = -x^2 + c, ]

where ( c ) is a constant to be determined.

Matching Coefficients:

• The leading term of ( s(x) ) is ( -x^{2n+2} ).

• The leading term of ( m(x)h(x) ) is ( (x^{2n})(-x^2) = -x^{2n+2} ), ensuring the degrees match.

• Comparing the constant term of ( s(x) ) using ( x = 0 ):

[ s(0) = p^(0) - 0 = 1, \text{ where } p^(x) = x^{2n} p\left( \frac{1}{x} \right). ]

• Evaluating ( m(0) ) and ( h(0) ):

[ m(0) = (-1)^n(n!)^2. ]

[ h(0) = -0 + c = c. ]

• Since ( s(0) = m(0)h(0) ):

[ 1 = (-1)^n (n!)^2 c, ]

which gives:

[ c = \frac{1}{(-1)^n(n!)^2} = \frac{(-1)^n}{(n!)^2}. ]

• For ( n ) even (since ( n ) is even), ( (-1)^n = 1 ), so ( c > 0 ).

Finding Additional Solutions:

Solving ( h(x) = 0 ):

[ -x^2 + c = 0 \Rightarrow x^2 = c \Rightarrow x = \pm \frac{1}{n!}, ]

since ( c = \frac{1}{(n!)^2} ).

Conclusion:

The additional real numbers ( x ) satisfying ( p\left( \frac{1}{x} \right) = x^2 ) are ( x = \pm \frac{1}{n!} ).