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I arrived at this question on reading about another issue, here.

I am unable to prove the above, even though attempt geometrical proof below:
Take $a,b,c,d$ as magnitudes of quantities (lengths). As no product term of $ab, cd$ is given so assuming right angle triangles (with right angle between sides of lengths $a,b$ & $c,d$), then $(a^2+b^2)$ leads to square of hypotenuse, let $h_1 (=\sqrt{a^2+b^2})$, similarly $(c^2+d^2)$ too, i.e. $h_2^2$. Their multiplication is $h_1^2h_2^2$.

The r.h.s. gives twice the product of sides of different triangles, i.e. $(a,c)$ & $(b,d)$. This algebraically means that the two triangles' sides should be multiplied. This in the simplest case is of a rectangle with sides of lengths $a,b,c,d$ with length $a=c, b=d$. This leads to r.h.s. as $2(a^2+b^2)$, so it leads to twice the sum of two hypotenuse's (let, $h_a, h_b$) square, i.e. $2(h_a^2+h_b^2)$. It satisfies for rectangle with side lengths $a=c, b=d$ & also for square with $a=b=c=d$. But, except for this simplest case, it makes difficult to interpret something useful from multiplying the different triangles sides.

I request proofs using any / all of the 3 criteria:
(i) geometrical,
(ii) algebraic,
(iii) complex number based approach

Update : Sorry, for faulty question, as one comment to post has shown.

jiten
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    Try $a=1,b=0,c=0,d=1$ – saulspatz Jul 04 '18 at 21:53
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    What are $a,b,c,$ and $d$? Are they related in some way that you haven't told us? Why would you assume there are right triangles involved, and why should we? – Cameron Buie Jul 04 '18 at 21:53
  • Maybe what you want is this. –  Jul 04 '18 at 21:58
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    Maybe the objective is to find a criterion that makes this true? – Mason Jul 04 '18 at 22:03
  • @Mason Please see the first line of edited O.P. – jiten Jul 04 '18 at 22:05
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    Well on the linked page it has written $a^2+b^2=c^2+d^2=2ac+2bd$ in a few spots... But you have written a product above. For the product we use this Fermat's Identity: $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$. – Mason Jul 04 '18 at 22:09
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    Let's delete this question and we'll meet you on https://math.stackexchange.com/questions/105330/equilateral-triangle-whose-vertices-are-lattice-points/105387#105387 If you post a question in the comments about how the argument goes through someone will surely pick up this discussion. – Mason Jul 04 '18 at 22:15
  • @Mason Have put a comment for the O.P. there, please guide why am getting wrong equality. – jiten Jul 04 '18 at 22:39
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    Done! Let me know if that helps. But you can do so there. As a comment – Mason Jul 04 '18 at 23:18

2 Answers2

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I'm sorry, but the claim is not true. Let $a = c = 1$ and $b = d = 2$. Then $(1+4)^2 \neq 2(1+4)$.

On the bright side, this explains why you were having trouble proving the claim.

HermitianCrustacean
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This

$$(a^2+b^2)(c^2+d^2)= 2ac + 2bd$$

is not true for all values of $a,b,c,d.$

As a counter example let $a=2,b=3,c=4,d=5$

We get $(13)(41)=16+30$ which is not true.

Mohammad Riazi-Kermani
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