Question

The sum of the squares of two consecutive integers is 41. find the integers

Answer 1

≡ We know that:
⇔ Consider the first integer is $a$
⇔ So, the next integer is $(a+1)$

≡ Solution:
⇒ $(a)^{2}+(a+1)^{2}=41$
⇒ $a^{2}+(a^{2}+2a+1)=41$
⇒ $2a^{2}+2a-40=0$
⇒ $(2a-8)(a+5)$
⇒ $a_{1}=\frac{8}{2}=\boxed{4}$║$a_{2}=-5$

∴ So, there are 2 options, $\boxed{4}and\boxed{5}$ or $\boxed{-5}and\boxed{-4}$

Answer 2

Start by declaring variables.
Since we have two consecutive integers, let your variables be x and x + 1.

Thus, we can say that:
x² + (x + 1)² = 41
x² + x² + 2x + 1 = 41
2x² + 2x - 40 = 0
x² + x - 20 = 0
(x + 5)(x - 4) = 0

So, x = -5 or 4.

Thus, we can have -5 and -4,
or 4 and 5.