Algebraic Number Problems

\(n^2-n=n(n-1)\). This is the product of two consecutive integers, one of which is even. Hence the product is divisible by \(2\).

\(n^3-n=n(n-1)(n+1)\). Among three consecutive integers, one is divisible by \(3\), and one is divisible by \(2\). Therefore the product is divisible by \(6\).

\(x^2-5x+6=(x-2)(x-3)\). Hence the roots are \(x=2\) and \(x=3\).

The possible integer roots are \(\pm1,\pm2\). Checking: \(1-4+2=-1\), \(-1+4+2=5\), \(8-8+2=2\), \(-8+8+2=2\). None is zero, so there are no integer roots.

The positive divisors of \(12\) are \(1,2,3,4,6,12\). Thus the pairs are \((1,12),(2,6),(3,4),(4,3),(6,2),(12,1)\).

It is enough to check \(n\equiv0,1,2,3,4\pmod5\). We get \(n^5\equiv n\pmod5\) for each residue: \(0^5=0\), \(1^5=1\), \(2^5=32\equiv2\), \(3^5=243\equiv3\), \(4^5\equiv(-1)^5\equiv-1\equiv4\). Hence \(n^5-n\equiv0\pmod5\).

Let \(u=x-y\), \(v=x+y\). Then \(uv=15\), \(x=\frac{u+v}{2}\), \(y=\frac{v-u}{2}\). Both factors are odd, so all divisor pairs give integer \(x,y\).

From \((u,v)=(1,15),(15,1),(3,5),(5,3)\) and the corresponding negative pairs, we get \((8,7),(8,-7),(4,1),(4,-1),(-8,-7),(-8,7),(-4,-1),(-4,1)\).

The numbers \(x,y\) are roots of \(t^2-10t+21=0\). The factorization \(t^2-10t+21=(t-3)(t-7)\) gives \(\{x,y\}=\{3,7\}\).

A square modulo \(3\) is \(0\) or \(1\). Then \(x^2+1\) has residue \(1\) or \(2\), while \(3y\equiv0\pmod3\). Contradiction.

Let \(u=x-2y\), \(v=x+2y\). Then \(uv=12\), \(x=\frac{u+v}{2}\), \(y=\frac{v-u}{4}\). We need \(u+v\) divisible by \(2\), and \(v-u\) by \(4\). Checking divisors of \(12\) gives \((u,v)=(2,6),(6,2),(-2,-6),(-6,-2)\). Hence \((x,y)=(4,1),(4,-1),(-4,-1),(-4,1)\).

\(n^3+3n^2+2n=n(n+1)(n+2)\). This is the product of three consecutive integers, so it is divisible by \(6\).

We get \((2x+y)^2+3y^2=28\). Thus \(3y^2\le28\), so \(y\in\{-3,-2,-1,0,1,2,3\}\). Checking these seven values gives all solutions: \((-3,1),(-3,2),(-2,-1),(-2,3),(-1,-2),(-1,3),(1,-3),(1,2),(2,-3),(2,1),(3,-2),(3,-1)\).

List integer factor pairs \(rs=12\): \((1,12),(2,6),(3,4)\) and permutations, plus negative pairs. The possible sums are \(13,8,7,-13,-8,-7\). Therefore \(a=-13,-8,-7,7,8,13\).

We get \(6x+6y=xy\), so \(xy-6x-6y=0\). Thus \((x-6)(y-6)=36\). Checking positive divisors of \(36\), we obtain \((x,y)=(7,42),(8,24),(9,18),(10,15),(12,12)\) and the symmetric pairs.

If \(p\nmid a\), then \(a\) has an inverse modulo \(p\). From \(a^2\equiv0\pmod p\), multiplying by the inverse of \(a\) twice gives \(1\equiv0\pmod p\), impossible. Hence \(p\mid a\).

Let the roots be \(r\) and \(s\). Then \(r+s=6\), \(k=rs\). Write \(r=3+t\), \(s=3-t\), where \(t\in\mathbb Z\). Then \(k=9-t^2\). Conversely, for every integer \(t\), the roots \(3+t\) and \(3-t\) are integers. Answer: all \(k=9-t^2\), \(t\in\mathbb Z\).

The equation is equivalent to \((x-y)^2=1\). Hence \(x-y=1\) or \(x-y=-1\). All solutions are \((x,y)=(t+1,t)\) and \((t,t+1)\), where \(t\in\mathbb Z\).

A square modulo \(4\) is \(0\) or \(1\). Thus the sum of two squares modulo \(4\) can be \(0,1,2\), but the right-hand side \(4z+3\) has residue \(3\). Contradiction.

We have \((2x-y)^2+3y^2=4\). Thus \(y\in\{-1,0,1\}\). If \(y=0\), then \(x=\pm1\). If \(y=1\), then \((2x-1)^2=1\), so \(x=0\) or \(1\). If \(y=-1\), then \((2x+1)^2=1\), so \(x=0\) or \(-1\). Answer: \((1,0),(-1,0),(0,1),(1,1),(0,-1),(-1,-1)\).

Multiply by \(2\): \((x-y)^2+(y-z)^2+(z-x)^2=4\). Squares of integers sum to \(4\). The only possible square patterns are \(4,0,0\) or \(1,1,2\), but \(2\) is not a square. Hence two differences are \(0\), and the third is \(\pm2\).

Thus two variables are equal, and the third differs from them by \(2\). All solutions are permutations of \((t,t,t+2)\) and \((t,t,t-2)\), where \(t\in\mathbb Z\).

Let \(\left(\frac{p}{q}\right)^n+a_{n-1}\left(\frac{p}{q}\right)^{n-1}+\cdots+a_0=0\). Multiplying by \(q^n\), \(p^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n=0\).

All terms except \(p^n\) are divisible by \(q\), so \(q\mid p^n\). Since \((p,q)=1\), we get \(q=1\).

\((x+y)^2=25+24=49\), so \(x+y=\pm7\). By Vieta, \(s=x+y\), hence \(s=7\) or \(s=-7\). Both values are attainable: root pairs \((3,4)\) and \((-3,-4)\).

Multiply by \(4\): \(4x^2-12x+4y^2-12y=0\). Then \((2x-3)^2+(2y-3)^2=18\).

The only odd squares summing to \(18\) are \(9\) and \(9\). Hence \(2x-3=\pm3\), \(2y-3=\pm3\). Thus \(x,y\in\{0,3\}\). All solutions are \((0,0),(0,3),(3,0),(3,3)\).

Assume a positive solution exists and choose one with minimal sum \(x+y\). Let \(x\ge y\). If \(x=y\), then \(2x^2=5x^2\), impossible. Thus \(x>y\).

Consider \(X^2-5yX+y^2=0\). One root is \(x\), the other by Vieta is \(x'=5y-x\), and the product of the roots is \(y^2\). Hence \(x'=\frac{y^2}{x}\), so \(x'\) is positive and integral.

Since \(x>y\), \(0