STEP I 2007 6


x^2-y^2=(x-y)(x+y)=(x-y)^3 \iff x+y=(x-y)^2=d^2

so combining with x-y=d we get

x+y+x-y=d^2+d \iff x=\frac{d(d+1)}{2}


x+y-(x-y)=d^2-d \iff y=\frac{d(d-1)}{2}.

setting x=\sqrt m,y=\sqrt n gives m-n=(\sqrt m -\sqrt n)^3 so by above

m=\left(\frac{d(d+1)}{2}\right)^2=\frac{d^2(d+1)^2}{4} \\n=\left(\frac{d(d-1)}{2}\right)^2=\frac{d^2(d-1)^2}{4} \\\begin{align} \\ m>n>10&\iff \frac{d^2(d+1)^2}{4}>\frac{d^2(d-1)^2}{4}>100 \\&\iff \frac{d(d+1)}{2}>\frac{d(d-1)}{2}>10 \\&\iff d(d+1)>d(d-1)>20 \\&\iff d\geqslant 6 \end{align}

so d=6\implies m=(\frac 1 2 (6)(7))^2=21^2=441,n=(\frac 1 2 (6)(5))^2=15^2=225 works. (d=6 is arbitrary, just for low numbers, this will work for all d larger than 6 (i think))


x^3-y^3=(x-y)(x^2+xy+y^2)=(x-y)^4\iff x^2+xy+y^2=(x-y)^3=d^3 \\\begin{align} \therefore d^3-d^2&=x^2+xy+y^2-(x-y)^2 \\ &= x^2+xy+y^2-x^2+2xy-y^2 \\ &= 3xy \end{align}

as required. now

\begin{align} 3xy=3x(x-d)&\iff 3x^2-3xd=d^3-d^2 \\&\iff 3x^2-3dx+(d^2-d^3)=0 \\&\iff x=\frac{3d±\sqrt{(3d)^2-4(3)(d^2-d^3)}}{2(3)} \\&\iff x=\frac{d±\sqrt{d^2-\frac{4(d^2-d^3)}{3}}}{2} \\&\iff x=\frac{d±\sqrt{\frac{3d^2-4d^2+4d^3}{3}}}{2} \\&\iff x=\frac{d±\sqrt{\frac{d^2(4d-1)}{3}}}{2} \\&\iff 2x=d±d\sqrt{\frac{4d-1}{3}} \end{align}

from this we also get that


we need x,y\in\mathbb{Z^+} so we require \sqrt{\frac{4d-1}{3}}\in\mathbb{Z^+}\iff \frac{4d-1}{3}=k^2 for k\in\mathbb{N}. so 4d=3k^2+1\iff d\equiv 1\pmod 3 as 4\equiv 1\pmod 3 (and d being 0 or 2 modulo 3 wouldn’t lead to 4d being 1 modulo 3). so we can write

\begin{align} d=3p+1&\implies 4(3p+1)=3k^2+1 \\&\iff 12p+3=3k^2 \\&\iff 4p+1=k^2 \\&\iff k^2 \equiv 1\pmod 4. \end{align}

so we can only pick odd squares for k^2\implies k must be odd.
picking k=3 gives p=\frac 1 3 (d-1)=2\iff d=7 which is fine as 7\equiv 1 \pmod 3.
this further gives m=\frac 1 2 d(1±k)=\frac 1 2 (7(1±3))=14,-7,n=\frac 1 2 d(-1±k)=\frac 1 2 (-7(1±3))=7,-14.
m,n\in \mathbb{N}\implies m=14,n=7.

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to verify 14^3-7^3=2744-343=2401 and \sqrt[4] {2401}=7=14-7

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