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

Revision as of 17:14, 6 September 2011 by Negativebplusorminus (talk | contribs) (Created page with "Since <math>a>b>c>d</math>, <math>a+b>c+d</math>, so <math>a+b>512</math> and since <math>a>b</math>, <math>a>512</math>. If we were to have <math>a-b\geq2</math>, then <math>a^...")
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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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