AoPS Wiki talk:Problem of the Day/September 6, 2011

Since $a>b>c>d$ , $a+b>c+d$ , so $a+b>512$ and since $a>b$ , $a>512$ .

If we were to have $a-b\geq2$ , then $a^2-b^2=(a+b)(a-b)\geq (a+b)+(a+b)>a+b+c+d$ , which defies the first equation. Thus, $a-b=1$ . Since $(a+b)+(c+d)=(a+b)(a-b)+(c+d)(c-d),$ we must also have $c-d=1$ .

From there, we can simply test values. We know $a>512$ , so we test $a=513$ . It fails, since the maximum possible base, $(a,b,c,d)=(513,512,511,510)$ , has a sum of only 2046.

Testing $a=514$ , we see that $(a,b,c,d)=(514,513,511,510)$ works, so the minimum is $\boxed{a=514}$ .

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AoPS Wiki talk:Problem of the Day/September 6, 2011