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Four different positive integers $a, b, c, d$ are such that $a^2 + b^2 = c^2 + d^2$. What is the smallest possible value of $abcd$?

$$a^2 - c^2 = d^2 - b^2$$

$$(a-c)(a+c) = (d-b)(d+b)$$

$$(a-c)(a+c) - (d-b)(d+b) = 0$$

So there is one pair I see:

$$a -c = d - b, a + c = d + b$$

Suppose WLOG, $a > b > c > d > 0$

But that doesnt strike anything.

TonyK
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Lebes
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  • http://math.stackexchange.com/questions/153603/diophantine-equation-a2b2-c2d2/736164#736164 – individ Feb 15 '15 at 17:21
  • Find an example. To be represented as the sum of two squares in at least two essentially different ways, the number has to be a product $xy$ where $x$ and $y$ are sums of squares. Fool around until you find a candidate, then argue you have managed to get the least possible $abcd$. The argument will be algebra-light. – André Nicolas Feb 15 '15 at 17:28
  • @AndréNicolas, well obviously, one of them is one, so let $a = 2$. Let $c = 1$, RHS $= 3$. But trial and error seems hard for $b, d$.? – Lebes Feb 15 '15 at 17:34
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    If WLOG $a > b > c > d > 0$, wouldn't that guarantee the strict inequality $a^2+b^2>c^2+d^2$? – graydad Feb 15 '15 at 17:34
  • You should have stopped at your second equation: $(a-c)(a+c)=(d-b)(d+b)$. (Hint: $1 \times 15 = 3 \times 5$.) – TonyK Feb 15 '15 at 17:43
  • There's this one http://math.stackexchange.com/questions/1117884/four-different-positive-integers-a-b-c-and-d-are-such-that-a2-b2-c2?rq=1 –  Feb 15 '15 at 17:45
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    The smallest number that can be expressed as the sum of two different squares in two different ways is $65=1^2+8^2=4^2+7^2$. That gives product $(1)(8)(4)(7)$. Now showing you can't beat that is not entirely pleasant. – André Nicolas Feb 15 '15 at 17:45

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