Pairs of squares

We begin by listing the factor pairs of $10^4=10000$ . $\begin{aligned} 10000 =1 \cdot 10000 \\ =2 \cdot 5000 \\ =4 \cdot 2500 \\ =5 \cdot 2000 \\ =8 \cdot 1250 \\ =10 \cdot 1000 \\ =16 \cdot 625 \\ =20 \cdot 500 \\ =25 \cdot 400 \\ =40 \cdot 250 \\ =50 \cdot 200 \\ =80 \cdot 125 \\ =100 \cdot 100 \\ \end{aligned}$

Note that $a^2-b^2=(a+b)(a-b)$ . These two factors must be integer factor pairs of $10000$ . Adding these two factors produces $2a$ (and similarly subtracting the second from the first produces $2b$ ), so the parity of the two factors must be the same. It is easy to see that there are exactly 8 factor pairs listed above that have factors of the same parity. However, we must not consider $(100,100)$ , as that produces $b=0$ and we are only concerned with positive integers. Thus the number of desired pairs is $\boxed{7}$ .

$10000=a^2-b^2=(a+b)(a-b)$ , sum of the pairs of multiplication (a+b) and (a-b) = 2a = even

Finding all the multiplication pairs of 10000,we get 13 pairs:

However, 5 of the pairs ${1\times 10000, 5\times 2000, 16\times 625, 25\times 40, and 80\times 125}$ have sum of the pairs is odd number

and the ${100\times 100}$ will have $a=100 and b=0$ where 0 is not a positive integer.

Hence, the number of ordered pairs of positive integers $(a,b)$ are $13-6-1=7$

$10^{4} = a^{2}-b^{2}$ $=> a^{2} = 100^2+b^{2}$

So we have to find all possible Pythagorean triplets $(a,100,b)$ where one side is known.

Now, $100$ has factors = $100, 50, 25, 20, 10, 5, 2, 1$ , each of which has to be used as the "known side" in Euclid's formula, which has been discussed here .

By Euclid's formula for Pythagorean triples, if one of the known sides is 100, we have the following triples,

$(2501,100,2499),(629,100,621),(125,100,75)$

If the known side is $50$ , we end up with the only triple $(626,50,624)$ . Multiplying throughout by $2$ , we get another triple $(1252,100,1248)$ .

For 'known side' $= 25$ , we have $(313,25,312)$ . Multiplying through out by $4$ , we have $(1252,100,1248)$ , which has already been counted earlier.

Proceeding with 'known side' $= 20$ , we have $(101,20,99)$ and $(29,20,21)$ . Multiplying throughout by $5$ , we have $(505,100,495)$ and $(145,100,105)$ .

For 'known side' $= 10$ , we get $(26,10,24)$ . Multiplying throughout by $10$ , we have $(260,100,240)$ .

For 'known side' $= 5$ and $2$ , we will get some triples which we already have got from other factors, so we will not count them. Finally, we arrive at the conclusion that the number unique triples is $7$

a^2 - b^2=(a+b)(a-b)... now we are assuming a-b as x and a+b as y... here we need to understand that both x and y have to be even nos.. we can write 10^4=2^4 * 5^4, so total no of factors of 10^4 = (4+1)(4+1)=25 so we can take 12 pair of (x, y) such that x*y=10^4, (where x and y are distinct), but here we need pairs of even nos.. so we have to reject (1,2^4 * 5^4),(5,2^4 * 5^3)....... and so on.. so the no of pairs which have an odd no is 5. So the ans is 12-5=7.