Math Olympiad | How to solve for "a²+b²" in this problem?

[@gmdFrame]

Hello, this is a good video! And you sound like a good person

[@RexxSchneider]

Sometimes "guess-and-check" and a bit of insight can be quicker than an analytic method. Guessing that an integer solution is available, we know that integers squared are mostly bigger than the integer, so we are probably going to find a solution when a^2 is a bit bigger than 43. Try a = ±7, which makes b = 49 - 43 = 6. Then b^2 = 36, making a = 36 - 43 = -7. Yup, that is a solution. So a^2 + b^2 = 49 + 36 = 85.

[@socute9248]

This seems way more complicated than necessary? It's simply a substitution problem Step 1: Set equations equal to find b, in terms of a a2-b=b2-a a2-b2=b-a (rearrange terms) (a+b)(a-b)=b-a (left side is difference of squares) a+b=-1 (after dividing both sides by a-b, which isn't 0 because it's given that a/=b) b=-a-1 Step 2: Substitute b=-a-1 in the first given equation and solve the quadratic. a2-(-a-1)=43 (substituting b=-a-1) a2+a-42=0 (arithmetic, shuffle terms) (a+7)(a-6)=0 (factor) Thus a=-7 or a=6 Step 3: You now have a-=7, b=6 (or vice versa), thus you have the answer: a2+b2=49+36=85

[@InDstructR]

I don't see how your way is much different honestly 😂 Step 1 is basically the same and for me step 2 would be slightly slower

[@socute9248]

​@@InDstructR I think the two ways are different! Your way is using linear combination, and the approach is to find a^2+b^2 as one expression -- without finding a and b independently. My way is using substitution to find a directly, then plugging it in to find a^2+b^2. In general I have found that students find substitution easier than linear combination, especially when the trick of how to multiply/add/subtract equations isn't obvious.

[@InDstructR]

​@@socute9248 yeah that definitely makes sense depending on the student