Determine all polynomials with integral coefficients $P(x)$ such that if $a,b,c$ are the sides of a right-angled triangle, then $P(a), P(b), P(c)$ are also the sides of a right-angled triangle. (Sides of a triangle are necessarily positive. Note that it's not necessary for the order of sides to be preserved; if $c$ is the hypotenuse of the first triangle, it's not necessary that $P(c)$ is the hypotenuse of the second triangle, and similar with the others.)
1. **Assume \( P(x) \) is a polynomial with integer coefficients.** We need to determine the form of \( P(x) \) such that if \( a, b, c \) are the sides of a right-angled triangle, then \( P(a), P(b), P(c) \) are also the sides of a right-angled triangle.

2. **Consider the sides \( x, x, \sqrt{2}x \) of a right-angled triangle.** These sides satisfy the Pythagorean theorem: 
   \[
   x^2 + x^2 = (\sqrt{2}x)^2
   \]
   We need \( P(x), P(x), P(\sqrt{2}x) \) to also form a right-angled triangle. This implies:
   \[
   2P(x)^2 = P(\sqrt{2}x)^2 \quad \text{or} \quad P(x)^2 + P(\sqrt{2}x)^2 = P(x)^2
   \]

3. **Analyze the equation \( 2P(x)^2 = P(\sqrt{2}x)^2 \).** This must hold for all \( x \). If \( P(x) \) is a polynomial of degree \( n \), let:
   \[
   P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0
   \]
   Substituting into the equation, we get:
   \[
   2(a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0)^2 = (a_n (\sqrt{2}x)^n + a_{n-1} (\sqrt{2}x)^{n-1} + \cdots + a_0)^2
   \]

4. **Simplify the right-hand side:**
   \[
   (a_n (\sqrt{2}x)^n + a_{n-1} (\sqrt{2}x)^{n-1} + \cdots + a_0)^2 = (a_n (\sqrt{2})^n x^n + a_{n-1} (\sqrt{2})^{n-1} x^{n-1} + \cdots + a_0)^2
   \]

5. **Equate the coefficients of \( x \) on both sides.** For the equation to hold for all \( x \), the coefficients of corresponding powers of \( x \) must be equal. This implies:
   \[
   2(a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0)^2 = (a_n (\sqrt{2})^n x^n + a_{n-1} (\sqrt{2})^{n-1} x^{n-1} + \cdots + a_0)^2
   \]

6. **Consider the leading term.** The leading term on the left-hand side is \( 2a_n^2 x^{2n} \). The leading term on the right-hand side is \( (a_n (\sqrt{2})^n x^n)^2 = a_n^2 2^n x^{2n} \). Equating these, we get:
   \[
   2a_n^2 = a_n^2 2^n
   \]
   For this to hold, \( a_n \neq 0 \) and:
   \[
   2 = 2^n \implies n = 1
   \]

7. **Conclude that \( P(x) \) is a linear polynomial.** Since \( n = 1 \), \( P(x) \) must be of the form:
   \[
   P(x) = a_1 x + a_0
   \]

8. **Check the constant term.** If \( P(x) = a_1 x + a_0 \), substituting into the original condition, we get:
   \[
   2(a_1 x + a_0)^2 = (a_1 \sqrt{2} x + a_0)^2
   \]
   Expanding both sides:
   \[
   2(a_1^2 x^2 + 2a_1 a_0 x + a_0^2) = a_1^2 2 x^2 + 2a_1 a_0 \sqrt{2} x + a_0^2
   \]
   Simplifying, we get:
   \[
   2a_1^2 x^2 + 4a_1 a_0 x + 2a_0^2 = 2a_1^2 x^2 + 2a_1 a_0 \sqrt{2} x + a_0^2
   \]
   Equating coefficients, we get:
   \[
   4a_1 a_0 x = 2a_1 a_0 \sqrt{2} x \implies 4a_1 a_0 = 2a_1 a_0 \sqrt{2}
   \]
   For this to hold, \( a_1 a_0 \neq 0 \) and:
   \[
   4 = 2\sqrt{2} \implies \sqrt{2} = 2 \quad \text{(contradiction)}
   \]
   Therefore, \( a_0 = 0 \).

9. **Conclude that \( P(x) = nx \) for some integer \( n \).** This satisfies the condition that \( P(a), P(b), P(c) \) are the sides of a right-angled triangle if \( a, b, c \) are the sides of a right-angled triangle.

The final answer is \( \boxed{ P(x) = nx } \) for any integer \( n \).