2021 JMPSC Accuracy Problems/Problem 13: Difference between revisions
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==Solution 2== | ==Solution 2== | ||
Note we are dealing with Pythagorean triples, so <math>xy=\{0,3,4 \}</math>, and we have <math>x+y</math> is a member of the set too. We see <math>x+y=4</math> has <math>x=\{1,3 \}</math> work, but <math>x+y=4</math> has nothing work. If <math>x+y=0</math>, we have <math>x=\{0,5 \}</math> work. The answer is <math>0+1+3+5=\boxed{9}</math> | Note we are dealing with Pythagorean triples, so <math>xy=\{0,3,4,5\}</math>, and we have <math>x+y</math> is a member of the set too. We see <math>x+y=4</math> has <math>x=\{1,3 \}</math> work, but <math>x+y=4</math> has nothing work. If <math>x+y=0</math>, we have <math>x=\{0,5 \}</math> work. The answer is <math>0+1+3+5=\boxed{9}</math> | ||
<math>\linebreak</math> | <math>\linebreak</math> | ||
~Geometry285 | ~Geometry285 | ||
Revision as of 11:33, 11 July 2021
Problem
Let 
and 
be nonnegative integers such that 
Find the sum of all possible values of 
Solution
Notice that since and are both integers, 
and 
are also both integers. We can then use casework to determine all possible values of :
Case 1: 
.
The solutions for and are the roots of 
, which are not real.
Case 2: 
.
The solutions for and are the roots of 
, which are not real.
Case 3: 
.
The solutions for and are the roots of 
, which are 
and 
.
Case 4: 
The solutions for and are the roots of 
, which are 
and 
.
Therefore, the answer is 
.
~kante314
~Revised and Edited by Mathdreams
Solution 2
Note we are dealing with Pythagorean triples, so 
, and we have is a member of the set too. We see 
has 
work, but has nothing work. If 
, we have 
work. The answer is 
~Geometry285
