I was recently preparing for the SOF IMO Level 2 exam. While practising for the exam, I encountered this question in my textbook about the chapter Square and Square Roots which also mentions the Pythagorean Triplets. This is the question:
$$22^2 + (m^2 - 1)^2 = 122^2$$
where we have to find the value of m. We could do it manually but there is an optimised solution given in the book I was unable to understand. This is exactly what the book says:
We have, $22^2 + (m^2 - 1)^2 = 122^2$
or $(11 \times 2)^2 + (m^2 - 1)^2 = (11^2 +1)^2$
Since, $2m$, $m^2 - 1$ and $m^2 + 1$ forms a Pythagorean Triplet,
So, we have m = 11
If the question were not having squares on every term, it would indeed be solvable via the triplet formula. But, since the square was added to every term, this does not feel right.
Is there anything I am missing?
Note that no matter what $m$ is, it is always true that $$(2m)^2 + (m^2-1)^2 = \underbrace{4m^2}_{(2m)^2} + \underbrace{m^4 -2m^2 + 1}_{(m^2-1)^2}\tag{clear parentheses}$$ $$=m^4 + 2m^2 + 1\tag{combine like terms}$$ $$ = (m^2 + 1)^2\tag{recognize perfect square trinomial}$$ That is, for any $m$ it is true that $(2m)^2 + (m^2-1)^2=(m^2+1)^2$.
Now, saying $$``a, b, \textrm{ and } c \textrm{ form a Pythagorean triple" (note no squares)}$$ means exactly $$``a^2 + b^2 = c^2" \;\;\; \textrm{(note squares).}$$
The "shortcut" is just saying that you can notice that taking $m=11$ produces the correct value for the $(2m)^2$ and $(m^2+1)^2$ terms, so the third term must match as well: $$(m^2 - 1)^2 = (11^2 - 1)^2$$ from which you may "observe" that $m=11$.
Note that this assumes $m$ is a nonnegative real number. There are other solutions if that assumption isn't made: $$(m^2 - 1)^2 = 120^2$$ $$m^2 - 1 = \pm 120$$ $$m^2 = 1 \pm 120$$ $$m^2 = 121 \;\;\; \textrm{or}\;\;\; m^2 = -119$$ $$m = 11, \;\;\; \textrm{or}\;\;\; m = -11, \;\;\; \textrm{or}\;\;\; m = i\sqrt{119}, \;\;\; \textrm{or}\;\;\; m = -i\sqrt{119}$$