Find all polynomials $ p$ of one variable with integer coefficients such that if $ a$ and $ b$ are natural numbers such that $ a \plus{} b$ is a perfect square, then $ p\left(a\right) \plus{} p\left(b\right)$ is also a perfect square.
1. **Preliminary Note**: We exclude the constant solutions \( P(x) = 2u^2 \) for some integer \( u \). Henceforth, the degree of \( P(x) \) is assumed to be \(\ge 1\).

2. **Lemma 1**: Let \( a \) be a given natural number which is not a perfect square. Then there exist infinitely many primes \( p \) for which \( a \) is a quadratic non-residue.

   **Proof**: By Euler's theorem, \(\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right) \cdot \left(\frac{b}{p}\right)\). We may assume that \( a \) is square-free by taking out the largest square divisor of \( a \). We can write \( a = 2^e \cdot p_1 \cdot \dots \cdot p_k \) where \( p_i \ge 3 \) are primes and \( e \in \{0,1\} \). If the product is empty, it shall be assumed to be equal to 1. If \( e = 0 \), then let \( p \equiv 1 \pmod{4p_1 \cdot \dots \cdot p_{k-1}} \) and \( p \equiv s \pmod{p_k} \) where \( s \) is a quadratic non-residue modulo \( p_k \). If \( e = 1 \), then let \( p \equiv 5 \pmod{8} \) and \( p \equiv 1 \pmod{p_1 \cdot \dots \cdot p_k} \). In any case, we see that 
   \[
   1 = \left(\frac{p}{p_i}\right) = (-1)^{\frac{p-1}{2} \cdot \frac{p_i-1}{2}} \cdot \left(\frac{p_i}{p}\right) = \left(\frac{p_i}{p}\right)
   \]
   for \( i = 1, \dots, k \). In the former case, however, the equality for \( i = k \) is changed to \(-1\) by our choice of \( s \) and in the latter, as \(\left(\frac{2}{p}\right) = (-1)^{\frac{p^2-1}{8}} = -1\); we conclude that \( a \) is a quadratic non-residue modulo \( p \) by using Gauss' Quadratic Reciprocity Theorem. The existence of infinitely many such primes is guaranteed by the Chinese Remainder Theorem and Dirichlet's Theorem on distributions of primes in an arithmetic progression. The proof is completed.

3. **Lemma 2**: Let \( f \) be a non-constant polynomial with integer coefficients such that \(\sqrt{f(n)}\) is an integer for all positive integers \( n \). Then there exists a polynomial \( g(x) \) with integer coefficients such that \( f(x) = g(x)^2 \).

   **Proof**: Let us write \( f(x) \) as a product of its irreducible divisors. Clearly, we can write 
   \[
   f(x) = p_1(x) \cdot \dots \cdot p_k(x) \cdot (R(x))^2
   \]
   where \( p_1, \dots, p_k \) are pairwise distinct irreducible divisors of \( f \) and all the polynomials \( p_1, \dots, p_k, R \) have integer coefficients. If the product \( p_1 \cdot \dots \cdot p_k \) is a constant, then it must be a square and we may conclude. Else, there exists a non-constant polynomial in the set \(\{p_1(x), \dots, p_k(x)\}\), which, say, is \( p_1(x) \). Then, as \( p_1 \) being irreducible has no common factor with the product \( h = p_2 \cdot \dots \cdot p_k \) (which might as well be empty or constant) and \( p_1' \) (the derivative of \( p_1 \)), we are assured of the existence of polynomials \( A(x), B(x), C(x), D(x) \) with integer coefficients and integers \( N_0, N_1 \neq 0 \) such that 
   \[
   A(x) \cdot p_1(x) + B(x) \cdot h(x) = N_0
   \]
   and 
   \[
   C(x) \cdot p_1(x) + D(x) \cdot p_1'(x) = N_1
   \]
   for all \( x \) by Euclid's Lemma. By Schur's theorem, the set of all primes dividing at least one value taken by \( p_1 \) over \(\mathbb{Z}\) is infinite, hence, we may choose a prime number \( p > N = \max(|N_0|, |N_1|) \) and an integer \( r \) such that \( p \mid p_1(r) \). Since \( p \nmid N_0 \), we see that \( p \nmid h(r) \). Now, choose \( x_0 = r + ip \) where \( 0 \le i \le p-1 \) is an integer we shall choose later. By the Taylor's Expansion of polynomials, we have 
   \[
   p_1(x_0) \equiv p_1(r) + ip \cdot p_1'(r) \pmod{p^2}
   \]
   In particular, we let \( p_1(r) = p^{j_0}k_0 \) and we can select \( i \) such that \( p^2 \nmid p_1(x_0) \) since the polynomials \( p_1 \) and \( p_1' \) are also coprime \( p \mid p_1(x_0), p_1'(x_0) \) will yield \( p \mid N_1 \), which is false as \( p > N \). Thus, \( v_p(p_1(x_0)) = 1 \) and \( v_p(h(x_0)) = 0 \). This yields that \( v_p(f(x_0)) \) is an odd number, a contradiction to the fact that it is a square. The result holds.

4. **Main Idea**: Notice that if \( a + b \) is a perfect square, then \( (ax^2 + bx^2) = (a + b)x^2 \) is also a perfect square. Define the polynomial \( T(x) = P(x^2) + P((t^2-1)x^2) \) where \( t \) is fixed for the moment. Then, \( T \) satisfies the conditions of Lemma 2 and we get that it is the square of a polynomial \( Q(x) \) with integer coefficients. Let \( d \) be the degree of \( P \), \( K \) be the leading coefficient of \( P \), and \( g(t) \) (not a polynomial/function etc) be the leading coefficient of \( Q \). The following equation is satisfied by comparing the leading coefficients in the equation 
   \[
   T(x) = Q(x)^2,
   \]
   \[
   K(1 + (t^2-1)^d) = g(t)^2
   \]
   for all \( t \) sufficiently large. Let \( K = mc^2 \) where \( m \) is a square-free number or \( m = 1 \). If \( m \) is square-free and \( m > 1 \), then we may write 
   \[
   (t^2-1)^d = m \cdot \left(\frac{g(t)}{mc}\right)^2 - 1
   \]
   where, for obvious reasons of divisibility, 
   \[
   \left(\frac{g(t)}{mc}\right)
   \]
   is an integer. Let \( p \) be a prime number and set \( t \equiv 1 \pmod{p} \), i.e., \( p \mid t^2 - 1 \) and so \( p \mid m \cdot \left(\frac{g(t)}{mc}\right) - 1 \) giving that \( m \) is a quadratic residue modulo \( p \). However, by Lemma 1, we can choose a sufficiently large prime \( p \) with \(\left(\frac{m}{p}\right) = -1\) and we get the requested contradiction. If \( m = 1 \), then suppose \( d > 1 \). If \( d \) is even, then \( 1 \) is a difference of sufficiently big perfect squares, and if \( d \) is odd, then we have 
   \[
   \left((t^2-1)^{\frac{d-1}{2}} - 1\right)^2 < \left(\frac{g(t)}{tc}\right)^2 < \left((t^2-1)^{\frac{d-1}{2}}\right)^2
   \]
   which is again a contradiction! Thus, we must have \( d = 1 \). If \( d = 1 \), then we have \( P(x) = vx + w \) and if \( a = 1, b = t^2 - 1 \), then we see that the equation 
   \[
   z^2 - vt^2 = w
   \]
   is satisfied for some integer \( z \). If \( v \) is a perfect square, then we get that \( w = 0 \). If not, then it is a Pell type equation, so its solutions must grow exponentially, however, they grow only linearly, as \( t \) grows linearly, only to be multiplied by a constant (equal to the largest perfect square divisor of \( v \)) and we get a contradiction!

Thus, the only solutions to this are the polynomials \( P(x) = k^2x \) for some integer \( k \) or \( P(x) = 2u^2 \) for some fixed integer \( u \). These also satisfy our given equation and we may conclude.

The final answer is \( \boxed{ P(x) = k^2x } \) for some integer \( k \) or \( P(x) = 2u^2 \) for some fixed integer \( u \).