It's Olympiad cuz you have to solve it the hardest way humanly possible.

@MathSync

Ай бұрын

I'll remember this next time. 😆

Easy! The solution is 11, an integer. The equation adds two roots. But roots only add up to an integer if they are integers. So the question is: Find 2 square numbers with difference 11. Let's see: 1, 4, 9, 16, 25, 36! There you have it.

@ntomata0002

Ай бұрын

So, you mean the solution is 36, since 36 - 25 = 11

@rainergro4055

Ай бұрын

Yes! Okay, additionaly, the roots need to add up to 11. 5 + 6 does. The equation works with odd numbers since consecutive squares are separated by odd numbers. So if you replace 11 with 13 the equation still works and the solution is the next square after 36 which is 49. The equation works as long you use the same odd (!) number twice.

@MathSync

11 күн бұрын

Great observation! You’ve correctly noted that the sum of two square roots will be an integer only if they are square roots of perfect squares. In this case, finding two perfect squares that differ by 11 is indeed the key. To elaborate on your solution: We are looking for two perfect squares, let’s call them ( a^2 ) and ( b^2 ), such that: a2−b2=11 This can be factored into: (a+b)(a−b)=11 Since 11 is a prime number, the factors of 11 are 1 and 11. So, we can set: a+b=11 a−b=1 Adding these two equations, we get: 2a=12 a=6 Subtracting the second equation from the first, we get: 2b=10 b=5 Therefore, the two perfect squares are ( 6^2 = 36 ) and ( 5^2 = 25 ), which indeed have a difference of 11. Your method of listing the perfect squares is a quick and intuitive way to arrive at the answer. Nicely done! 👍