High School Can You Discover the Integer Pair That Satisfies This Unique Sum Equation?

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The problem involves finding a pair of positive integers \( (a, b) \) where \( a < b \) that satisfies the equation \( 37^2 + 46^2 + 49^2 - 20^2 - 17^2 = a^2 + b^2 \). Participants are encouraged to submit their solutions following the guidelines provided. Members castor28 and kaliprasad successfully solved the equation, showcasing their mathematical skills. The thread highlights the importance of collaboration and problem-solving within the community. Engaging with such challenges can enhance mathematical understanding and foster a supportive learning environment.
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Here is this week's POTW:

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Find a pair $(a,\,b)$ of positive integers $a<b$ such that

$37^2+46^2+49^2-20^2-17^2=a^2+b^2$

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to the following members for their correct solution:):

1. castor28
2. kaliprasad

Solution from castor28:
We may write:
\begin{align*}
S &= 37^2+46^2+49^2-20^2-17^2\\
&= 37^2 +(46^2 - 20^2) + (49^2 - 17^2)\\
&= 37^2 + (66\times26) + (66\times32)\\
&= 37^2 + 66\times58 \\
&= 37^2 + 66\times(66-8)\\
&= 37^2 + 66^2 - 8\times66\qquad [1]
\end{align*}
We notice that $8+66=2\times37$; this suggests to write $37^2 - 29^2 = 8\times66$; substitution in [1] gives $S = 29^2+66^2$, i.e., $a=29$, $b=66$.
 

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