# Integers $a,b$

How many pairs of integers $(a,b)$ satisfy $a^2-b^2=2000^2?$

Note by Sudipto Podder
3 months ago

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@Sudipto Podder, i have checked my formula for different values from wolframalpha.

- 2 months, 3 weeks ago

👍🏻

- 2 months, 3 weeks ago

Here is my process bro.

- 2 months, 4 weeks ago

Thanks

- 2 months, 4 weeks ago

Welcome!

- 2 months, 3 weeks ago

You can apply this formula on your question as well. I have seen that your thought process was similar.

- 2 months, 4 weeks ago

Look my thought was something like: a²-b²=2000² Or, (a+b)(a-b)= 2000² Now , we have to express 2000² like such k×m where we can say that (a+b)=k and (a-b)=m ...and when we use addition in those equations , we get 2a =k+m ....but then .. a can be an integer if (k+m) is an even number....in that case k and m should be in the same parity ...... Now 2000²=(2^8)×(5^6) Here both k and m can't be odd.... because there should be 2 as a factor in at least one of k and m........so both k and m should be even...now you can express (2^8)×(5^6) as k×m in total {(8+1)×(6+1)}=63 ways . But in both k and m, there should be both 2 and 5 as factors.... otherwise one of them will definitely be odd .....here in those 63 ways , there are {(6+1)×2}=14 ways where there are one odd and one even between k and m.....so total positive ways of expressing 2000² as k×m( following all the conditions) are of (63-14)=49 types...... But here you have to notice that all these discussions are for positive integers...but it is also possible for negative integers ...so the total ways are of (49×2)=98 types SO, THE ANSWER IS [{(98)}]

- 2 months, 4 weeks ago

Your intution is correct,but i think you don't know how to calculate even factors.See my answer.

- 2 months, 4 weeks ago