Factorisation Methods

Factor out \(3xy\): \(6x^3y-9x^2y^2+3xy^3=3xy(2x^2-3xy+y^2)\).

Then \(2x^2-3xy+y^2=(2x-y)(x-y)\). Hence the full factorisation is \(3xy(2x-y)(x-y)\).

We have \(ab+ac+bd+cd=a(b+c)+d(b+c)\).

Factoring out \(b+c\), we get \((a+d)(b+c)\).

The two numbers are \(p\) and \(q\). Therefore \(x^2-(p+q)x+pq=(x-p)(x-q)\).

Let \(U=2a+b\), \(V=a-2b\). Then \(U-V=a+3b\), \(U+V=3a-b\).

Thus the expression equals \((a+3b)(3a-b)\).

We have \(a^2+2ab+b^2-c^2=(a+b)^2-c^2\).

This is a difference of squares: \((a+b-c)(a+b+c)\).

\(x^4+4y^4=x^4+4x^2y^2+4y^4-4x^2y^2=(x^2+2y^2)^2-(2xy)^2\).

Hence \(x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2)\).

For odd \(m\), \(a^m+b^m=(a+b)(a^{m-1}-a^{m-2}b+\cdots-ab^{m-2}+b^{m-1})\).

With \(m=2n+1\), the factor \(a+b\) appears, so the divisibility is proved.

Let \(P(x)=x^3+kx^2-5x-2k\). The condition is equivalent to \(P(2)=0\).

\(P(2)=8+4k-10-2k=2k-2\), hence \(k=1\).

Then \(P(x)=x^3+x^2-5x-2=(x-2)(x^2+3x+1)\).

\(x^4+x^2+1=x^4+2x^2+1-x^2=(x^2+1)^2-x^2\).

Thus the factorisation is \((x^2-x+1)(x^2+x+1)\).

We have \(n^4+4n^2+3=(n^2+1)(n^2+3)\).

For positive \(n\), both factors are greater than \(1\), so the number is composite.

By the identity, \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\).

The second factor is an integer when \(a,b,c\) are integers. Therefore the expression is divisible by \(a+b+c\).

\(x^2y^2-9z^4=(xy)^2-(3z^2)^2\).

Hence \(x^2y^2-9z^4=(xy-3z^2)(xy+3z^2)\).

The condition gives \(a^3+b^3+c^3-3abc=0\). By the identity, this equals \((a+b+c)(a^2+b^2+c^2-ab-bc-ca)\).

Since \(a+b+c>0\), the second factor is zero. But \(2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2\). Hence \(a=b=c\).

We have \(x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2)\).

The first factor is \((x-y)^2+y^2\). For positive integers \(x,y\), it equals \(1\) only when \(y=1\) and \(x=y\), so \(x=y=1\).

Then the number is \(5\), which is prime. In all other cases both factors are greater than \(1\), so the number is composite. Answer: \((1,1)\).

\(x^6-1=(x^3-1)(x^3+1)\).

Also \(x^3-1=(x-1)(x^2+x+1)\), and \(x^3+1=(x+1)(x^2-x+1)\).

Thus \(x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1)\).

\(a^6-b^6=(a^3-b^3)(a^3+b^3)\).

Also \(a^3-b^3=(a-b)(a^2+ab+b^2)\). Therefore \(a^2+ab+b^2\) is indeed a factor of \(a^6-b^6\).

Rewrite as \(x^2-2x-y^2=6\). Then \((x-1)^2-y^2=7\).

Thus \((x-1-y)(x-1+y)=7\). The possible factor pairs are \((1,7)\), \((7,1)\), \((-1,-7)\), \((-7,-1)\).

Solving the systems gives \((x,y)=(5,3),(5,-3),(-3,-3),(-3,3)\).

\(a^4+a^2b^2+b^4=(a^2+b^2)^2-a^2b^2\).

This is a difference of squares: \((a^2-ab+b^2)(a^2+ab+b^2)\).

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

Since the degree is \(3\), \(E=k(a-b)(b-c)(c-a)\). Substitute \(a=1\), \(b=2\), \(c=3\): \(E=2\), and \((a-b)(b-c)(c-a)=2\). Hence \(k=1\).

Thus \(E=(a-b)(b-c)(c-a)\).

The expression is \((n+1)^4+4\). Put \(m=n+1\). Then \(m^4+4=(m^2-2m+2)(m^2+2m+2)\).

If \(|m|>1\), this is a non-trivial factorisation. If \(m=0\), the number is \(4\), not prime. If \(m=1\) or \(m=-1\), the number is \(5\), which is prime.

Hence \(n=0\) or \(n=-2\).

Assume \(n^4+4=m^2\). Then \(m^2-n^4=4\), so \((m-n^2)(m+n^2)=4\).

Both factors are positive. The possible pairs are \((1,4)\), \((2,2)\), \((4,1)\). The pair \((1,4)\) gives \(2m=5\), impossible. The pair \((2,2)\) gives \(n=0\), contradicting positivity. The pair \((4,1)\) is impossible by order.

Therefore \(n^4+4\) is not a square.

We have \((x-y)(x+y)=24\). Let \(u=x-y>0\), \(v=x+y\). Then \(uv=24\), and \(u\) and \(v\) have the same parity.

The suitable positive pairs are \((2,12)\) and \((4,6)\). From \(x=\frac{u+v}{2}\), \(y=\frac{v-u}{2}\), we get \((x,y)=(7,5)\), \((5,1)\).

If \(v<0\), then \(uv<0\) since \(u>0\), so there are no other solutions.

The left side vanishes when \(a=b\), when \(b=c\), and when \(c=a\). Hence it is divisible by \((a-b)(b-c)(c-a)\).

Since the degree is \(3\), the left side equals \(k(a-b)(b-c)(c-a)\). Substitute \(a=1\), \(b=2\), \(c=3\). The left side is \(-2\), while the product is \(2\). Thus \(k=-1\).

The identity is proved.

Let \(s_k=a^k+b^k+c^k\). Since \(a,b,c\) are roots of \(t^3+qt-r=0\), where \(q=ab+bc+ca\), \(r=abc\), each root satisfies \(t^3=-qt+r\).

Multiplying by \(t^2\) and summing over the three roots gives \(s_5=-q s_3+r s_2\).

When \(a+b+c=0\), we have \(s_3=3abc=3r\), and \(s_2=-2q\). Thus \(s_5=-3qr-2qr=-5qr\).

Since \(q=-\frac{s_2}{2}\), we obtain \(s_5=\frac{5}{2}rs_2=\frac{5}{2}abc(a^2+b^2+c^2)\).

Let the expression be \(E\). When \(a=b\), \(b=c\), or \(c=a\), it becomes zero, so \(E\) is divisible by \((a-b)(b-c)(c-a)\).

The remaining factor has degree \(2\) and is symmetric, so it has the form \(\alpha(a^2+b^2+c^2)+\beta(ab+bc+ca)\). Put \(c=0\). Then \(E=a^4b-b^4a=ab(a-b)(a^2+ab+b^2)\), while \((a-b)(b-c)(c-a)=-(ab)(a-b)\). Hence the remaining factor is \(-(a^2+ab+b^2)\) when \(c=0\).

Thus \(\alpha=-1\), \(\beta=-1\), and

\[E=-(a-b)(b-c)(c-a)(a^2+b^2+c^2+ab+bc+ca).\]

The obvious attempt to take all factors large is not helpful: the product tends to decrease. The hidden idea is that small factors may become negative, and two negative factors give a positive product.

Let the original numbers be \(1,1,a\). The original product is \(a\), while the new product is \((-3)(-3)(a-4)=9a-36\).

We need the new product to be \(2020\) greater:

\[9a-36=a+2020.\]

Thus \(8a=2056\), so \(a=257\). Check: the original product is \(1\cdot1\cdot257=257\), the new product is \((-3)(-3)\cdot253=2277\), and \(2277-257=2020\). Therefore it is possible.

The point is not to guess all five numbers, but to create a large coefficient in front of one unknown. Let the numbers be \(1,1,1,1,a\). The original product is \(a\).

After decreasing each factor by \(4\), the product becomes \((-3)^4(a-4)=81a-324\).

We require

\[81a-324=78a.\]

Hence \(3a=324\), so \(a=108\). Check: the original product is \(108\), the new product is \((-3)^4\cdot104=8424\), and \(78\cdot108=8424\). Therefore \(1,1,1,1,108\) works.

The task is not about brute force; it is about controlling the sign. If we take five ones and one three, then after decreasing by \(4\) we get the factors \(-3,-3,-3,-3,-3,-1\), whose product is positive and equals \(243\).

Let the original numbers be \(1,1,1,1,1,3,a\). The original product is \(3a\), while the new product is \(243(a-4)\).

We want an \(80\)-fold increase:

\[243(a-4)=80\cdot3a.\]

Thus \(243a-972=240a\), so \(a=324\). Check: the original product is \(972\), the new product is \(243\cdot320=77760\), and \(80\cdot972=77760\). Therefore such a set exists.