#1 Algebraic Expressions and Identities

Let \(U=x^2+2x+3\), \(V=x^2-3\). The expression equals \((U-V)(U+V)\).

We have \(U-V=2x+6=2(x+3)\), \(U+V=2x^2+2x=2x(x+1)\). Hence the factorization is \(4x(x+1)(x+3)\).

Use \(u^3+v^3=(u+v)(u^2-uv+v^2)\), with \(u=2a\), \(v=3b\).

Thus \(8a^3+27b^3=(2a+3b)(4a^2-6ab+9b^2)\).

Group: \[ a^2b-a^2c+b^2c-bc^2=a^2(b-c)+bc(b-c). \] Factor out \(b-c\):

\((b-c)(a^2+bc)\).

We have \(x^2+y^2=7^2-2\cdot10=29\).

Also \(x^3+y^3=(x+y)(x^2-xy+y^2)=7(29-10)=133\).

If \(n\) is odd, then \(n^3\) is odd, and \[ n^3=\left(\frac{n^3+1}{2}\right)^2-\left(\frac{n^3-1}{2}\right)^2. \]

If \(n\) is even, write \(n=2m\). Then \(n^3=8m^3\). Take \(A=2m^3+1\), \(B=2m^3-1\). Then \(A^2-B^2=(A-B)(A+B)=2\cdot4m^3=8m^3=n^3\).

Transform: \[ x^4+4y^4=x^4+4x^2y^2+4y^4-4x^2y^2=(x^2+2y^2)^2-(2xy)^2. \] Hence \[ x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2). \]

For \(x>1\), \(y\ge1\), both factors are greater than \(1\). Thus the number is composite.

By the sum of cubes identity, \[ x^3+y^3=(x+y)^3-3xy(x+y). \] Since \(x+y=1\), we get \(x^3+y^3=1-3xy\).

Thus \(x^3+y^3+3xy=1\).

From \((a+b+c)^2=0\), we get \(a^2+b^2+c^2+2ab+2bc+2ca=0\). Hence \(a^2+b^2+c^2=-2(ab+bc+ca)\).

Also \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\). Thus \(a^3+b^3+c^3=3abc\).

We have \(n^2-1=(n-1)(n+1)\). For \(n>2\), both factors are greater than \(1\): \(n-1>1\), \(n+1>1\).

Therefore \(n^2-1\) is a product of two integers greater than \(1\), so it is composite.

Squaring, \[ \left(x+\frac1x\right)^2=x^2+2+\frac1{x^2}=9. \] Hence \(x^2+\frac1{x^2}=7\).

Write \[ a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}). \] Expanding shows that all intermediate terms cancel.

Hence \(a^n-b^n\) has factor \(a-b\).

Let \(s=x+y=4\), \(p=xy\). Then \(x^3+y^3=s^3-3ps=64-12p\).

Since \(64-12p=28\), we get \(p=3\). Thus \(x,y\) are roots of \(t^2-4t+3=0\), so \(t=1\) or \(t=3\). The pairs are \((1,3)\), \((3,1)\).

By the identity, \[ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). \] The left side is zero, and \(a+b+c>0\), so the second factor is zero.

But \[ 2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2. \] A sum of squares is zero only when \(a=b=c\).

We have \(x^4+2x^2y^2+y^4=(x^2+y^2)^2\). Thus the expression is \((x^2+y^2)^2-4^2\).

By difference of squares, it factors as \((x^2+y^2-4)(x^2+y^2+4)\). Over the reals, the second factor does not split further.

From \(a+b+c=0\), \(a^2+b^2+c^2=-2(ab+bc+ca)\). Square: \[ (a^2+b^2+c^2)^2=4(ab+bc+ca)^2. \]

The left side is \(a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)\). The right side is \(4(a^2b^2+b^2c^2+c^2a^2+abc(a+b+c))\), and the last term is zero. This gives the required equality.

Use \(n^4+4=(n^2-2n+2)(n^2+2n+2)\).

If \(n>1\), both factors are greater than \(1\), so the number is composite. For \(n=1\), \(1^4+4=5\), which is prime. Answer: \(n=1\).

It is enough to prove \(a^2+b^2+c^2-ab-bc-ca\ge0\). Multiply by \(2\): \[ 2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2. \]

The right side is non-negative, so the inequality is proved.

Let the expression be \(E\). It has degree \(3\), and if \(a=b\), then \(E=0\), so \(a-b\) is a factor. Similarly, \(b-c\) and \(c-a\) are factors.

Since \((a-b)(b-c)(c-a)\) also has degree \(3\), only the sign remains. Substitute \(a=1\), \(b=2\), \(c=3\): \(E=1(-1)+4(2)+9(-1)=-2\). But \((a-b)(b-c)(c-a)=(-1)(-1)(2)=2\). Hence \(E=-(a-b)(b-c)(c-a)\).

Compute: \[ (x-1)^2+(y-1)^2+(z-1)^2=x^2+y^2+z^2-2(x+y+z)+3. \] By the conditions this is \(3-6+3=0\).

A sum of squares is zero only when each square is zero. Hence \(x=y=z=1\).

We have \((x-y)(x+y)=15\). The two factors have the same parity because their sum is \(2x\), even. Since \(15\) is odd, both are odd.

The possible pairs are \((1,15)\), \((3,5)\), \((-1,-15)\), \((-3,-5)\). Solving \(x-y=u\), \(x+y=v\), we get \((x,y)=(8,7),(4,1),(-8,-7),(-4,-1)\).

Since \(a+b+c=0\), the identity gives \(a^3+b^3+c^3=3abc\).

Given \(abc=2\), we get \(a^3+b^3+c^3=6\).

Consider the difference: \[ a^4+b^4-a^3b-ab^3=a^3(a-b)-b^3(a-b)=(a-b)(a^3-b^3). \] Also \(a^3-b^3=(a-b)(a^2+ab+b^2)\).

So the difference is \((a-b)^2(a^2+ab+b^2)\). The first factor is non-negative, and the second is \(\left(a+\frac b2\right)^2+\frac34b^2\ge0\). Hence the inequality is proved.

We have \(n^4+4n^2+4=(n^2+2)^2\). A square of an integer greater than \(1\) is never prime.

Here \(n^2+2\ge3\), so \((n^2+2)^2\) has a divisor \(n^2+2\), different from \(1\) and from the number itself. Hence there are no solutions.

Let \(s=x+y\), \(p=xy\). From \(x^2+y^2=s^2-2p=10\), we get \(p=\frac{s^2-10}{2}\).

Also \(x^3+y^3=s^3-3ps=10\). Substitute \(p\): \[ s^3-\frac{3s(s^2-10)}{2}=10. \] Multiplying by \(2\), \(2s^3-3s^3+30s=20\), so \(s^3-30s+20=0\) after changing sign.

Additionally, real \(x,y\) require \(s^2-4p=s^2-2(s^2-10)=20-s^2\ge0\). Thus the possible values of \(x+y\) are precisely the real roots of \(s^3-30s+20=0\) lying in \([-\sqrt{20},\sqrt{20}]\).

We have \[ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). \] Since \(a,b,c\ge0\), the first factor \(a+b+c\ge0\).

The second factor is non-negative because \[ 2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2. \] Hence the whole difference is non-negative. Equality occurs when \(a+b+c=0\), which gives \(a=b=c=0\), or when the second factor is zero, which gives \(a=b=c\). Thus equality holds exactly when \(a=b=c\).