Math Olympiad Question | Equation Solving | You should learn this trick!

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[@ligion324]

Let c=b+2, then a^2+c^2=100. Since it's symmetric, you only need to verify 5 cases (1,2,3,4,5) instead of 7 cases. You can also save the efforts of proving b

[@singank08]

All values of a between (-10,10) will give two solution for b between [-12,8] and a=±10 will give b=-2 as solution.

[@AdamGlesser]

The end feels a bit unnecessary. Once you know that a² + (b+2)² = 100, you simply observe that the only way to write 100 as a sum of positive squares is 8²+6² (this is quickly seen by exhaustion). Thus, (8, 6-2) and (6, 8-2) are the only solutions.

[@ihlin0733]

I was just about to leave the same comment.

[@Viki13]

It's also a pythagorean triplet a multiple of the famous 3,4,5 triplet

[@ligion324]

No. You observed it but you didn't prove it.

[@AdamGlesser]

@@ligion324 Actually, I did. As I said, it is a proof by exhaustion. 100-81 = 19 is not a square. 100-49 = 51 is not a square. There are no other cases that need to be checked (even the last case is strictly unnecessary) since all squares less than 50 would have to be paired with a square above 50.

[@ligion324]

@@AdamGlesser "all squares less than 50 would have to be paired with a square above 50". You didn't mention or hint this in your original comment. Without such statement you have to explore all possible cases but you didn't. That's why I said you didn't prove it.

[@alexl6671]

a^2 + (b+2)^2 = 100 if a >= b+2 then 7 < sqrt(50)

[@MichaelRothwell1]

As I solved the equation from the thumbnail, I didn't see the restriction to positive integers, so got more solutions, 12 in fact. a²+b²+4b+4=100 a²+(b+2)²=100=4×5² a and b are both even, as the square of an odd number is of the form 4n+1 (a/2)²+(b/2+1)²=5² As the only solutions in positive integers to x²+y²=5² are x=3, y=4 & vice versa, Solutions are |a/2|=0, |b/2+1|=5: a=0, b=8, -12 |a/2|=3, |b/2+1|=4: a=±6, b=6, -10 |a/2|=4, |b/2+1|=3: a=±8, b=4, -8 |a/2|=5, |b/2+1|=0: a=±10, b=-2