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Four different integers form an increasing A.P. One of these numbers is equal to the sum of the squares of the other three numbers. Find the numbers

My Attempt:

I assumed the four numbers as: $(a-3d), (a-d), (a+d), (a+3d)$ but now i can't understand which number to square.

Harshit Joshi
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  • There are only four possibilities. Why not simply try each one in turn? – Barry Cipra Apr 28 '18 at 12:52
  • Woudn't that make this problem way too lengthy. It's an objective problem so it needs to be solved quickly. – Harshit Joshi Apr 28 '18 at 12:55
  • Plus what if there were say a 100 numbers?? – Harshit Joshi Apr 28 '18 at 12:56
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    But there aren't a hundred numbers, there are only four. My point is, when stuck on a problem like this, do something. Even if it doesn't lead to a solution, playing around with the formulas gives the front of your brain something to do while the back of your brain is grinding away. (Another tactic is to look for "cheap" solutions with small numbers, say $(-1)^2+0^2+1^2$....) – Barry Cipra Apr 28 '18 at 13:12

1 Answers1

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Hint: Clearly the largest number must be the sum of the square of the other three. Otherwise, the square of the largest one would be in the sum, and the square of the largest one, plus other stuff, is larger than any of the other terms.

Arthur
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  • But which is the largest term. Note that a and d are not necessarily positive. – Harshit Joshi Apr 28 '18 at 12:55
  • @HarshitJoshi $d$ may be assumed positive (if it isn't, then just reverse the order of the numbers before you assign $a$ and $d$). And the sign of $a$ doesn't change anything: a negative number is smaller than a positive number. So the one with $+3d$ is necessarily largest. – Arthur Apr 28 '18 at 12:57
  • Okay so i get this relation $a+3d = 3a^2+11d^2-6ad$. How should i proceed now??? – Harshit Joshi Apr 28 '18 at 13:03
  • Okay sorry, it turns out this question is answered already. I will delete it. – Harshit Joshi Apr 28 '18 at 13:06
  • Now you proceed by noting that if $a$ or $d$ are too large (as in "larger than $2$"), then the right-hand side is larger than the left-hand side, because of the squares. This is easier to see if you write it as $3a^2+2d^2 +(a-3d)^2$. Then it's easy enough to check all cases. – Arthur Apr 28 '18 at 13:14