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My friend comes up to me and says that you have 3 minutes to solve this question:
What does $(7^2 + 5^2 )(6^2 + 11^2 )(3^2 + 13^2 )(7^2 + 8^2 )$ equals to:

A) $(86^2 + 76^2)(130^2 + 29^2)$
B) $(97^2 + 47^2)(125^2 + 67^2)$
C) $(103^2 + 78^2)(47^2 + 88^2)$
D) $(9^2 + 91^2)(125^2 + 111^2)$
(Multiple correct options may exist!)

Now I thought, and thought why not brute force! It should work right? But then the clock told me no time for that. I cant think of anything else, Now I need your help with what other methods could I use to do this quickly. All unique ideas are welcome! Thanks
(I'm not asking for exact solution to the problem. Maybe a hint or two or some examples about the method you would apply to solve this)

EDIT: Thanks for pointing it out I put an extra 2 by mistake in all integers given in the question. It has been corrected now

marks_404
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  • thos optioins look very small – Asinomás Apr 11 '21 at 19:48
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    The product on top is at least $(100^2)(100^2)(70^2)(80^2)$ and the one on the bottom is at most $(2\cdot(200)^2)(2\cdot(200)^2) = 64\cdot 100^4$ – Asinomás Apr 11 '21 at 19:50
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    @HereToRelax has a good observation. Are you sure you have this right? There is a standard identity for the product of the sums of squares based on complex numbers. There are various tests you can do (eg work modulo $9$ or $10$ to eliminate options). but the answers are way too small as they stand. – Mark Bennet Apr 11 '21 at 19:56
  • Please check the edit. I added an extra 2 at the end of all integers in the question by mistake – marks_404 Apr 11 '21 at 20:03
  • @saulspatz indeed and noting that $C$ is an odd number will eliminate it. – Mark Bennet Apr 11 '21 at 20:03
  • Do you know complex numbers? Or the identity for multiplying products of two squares to get an answer as a sum of two squares? – Mark Bennet Apr 11 '21 at 20:04
  • @MarkBennet I do know the basics of complex numbers however I am unaware of this identity it would be great if you could elaborate! – marks_404 Apr 11 '21 at 20:05
  • Try $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ - I could tell you the identity, but it will fix in your mind better if you work it out for yourself. – Mark Bennet Apr 11 '21 at 20:07
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    $$(a^2+b^2)(c^2+d^2)=(ac-db)^2+(ad+bc)^2$$ – Thomas Andrews Apr 11 '21 at 20:08
  • @MarkBennet Taking the modulus after that !? – marks_404 Apr 11 '21 at 20:09
  • Indeed, and see what @ThomasAndrews has put. That's how I recreate the identity if I need it. – Mark Bennet Apr 11 '21 at 20:10
  • Guys ... Depending on the problem, this complex number identity may help, or may not. (One might need to first break up e.g. $7^2+5^2=(1^2+1^2)(6^2+1^2)$ etc. and then group the resulting sums of squares into two chunks ...) I guess I would just flatly refuse to do it in $3$ minutes. Why? Because brute force is indeed the easiest and most straightforward way to solve this. $3$ min limit just serves as a test how quickly you can calculate, which is an irrelevant skill these days. Surely you may find some shortcuts when doing brute force, but it remains brute force all the same! –  Apr 11 '21 at 20:16
  • Actually there are two identities, because $$(a^2+b^2)(c^2+d^2)=(a+bi)(a-bi)(c+di)(c-di)\=(a+bi)(c+di)\overline{(a+bi)(c+di)}\=(a+bi)(c-di)(a-bi)(c+di)\=(a+bi)(c-di)\overline{(a+bi)(c-di)}$$ – saulspatz Apr 11 '21 at 20:17
  • A modulo (10) argument eliminates one of the choices. Imagine my surprise when I manually checked the other 3 choices (with a calculator). Hard to imagine solving this in one's head in under 3 minutes. For example, it doesn't seem reasonable that you would be able to calculate (for example) two 2-digit numbers in your head, and retain the sum as you did other calculations. – user2661923 Apr 11 '21 at 20:57

2 Answers2

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HINT:

Half of the answer $\to $ Quickly check the last digit of each product.For example, the original product ends with 2: (a4)(b7)(c8)(d3)=(e2). You can go over the 4 options in similar way within a minute to eliminate (C) and (D), giving you 2 minutes to choose between (A) and (B).

Star Bright
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Consider the complex number $Z = (7+5i)(7+8i)(6+11i)(3+13i)$

Simplifying, $Z=(49-40 +i(56+35))(18-143+i(78+33))$

$Z=(9+91i)(-125+111i)$

Compare the modulus now, you can work out similarly for the other possibilities.

I got the second expression for $Z$ by multiplying first and second independently from the product of third and fourth.

Aditya Sharma
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