since 2009 is 49*4. I used modulo 7 to arrive at: a^2 + b^2 = 0(mod7) if a or b is non zero then dividing by it makes the eq: 1 + u^2 = 0 (mod 7)

Such a square doesn't exist. ( just square 1 to 6 and check the non of them are -1 mod 7).

which implies both a and b are divisible by 7. Setting a=7x and b= 7y, we can reduce the equation to x^2 + y ^2 = 41.

Doing mod 4 shows that one is even and the other is odd.

Checking even numbers 0,2,4,6 = we find only one solution 16 + 25 =41

i.e x=5 and a = 35

Note that since a > b, it follows that $2a^2$ > $a^2 + b^2$ > 2009. This yields $a^2$ > 1004 and so 45 > a > 31(For if a > 45 then $a^2 > 2025$ . Also looking at the equation modulo 5, $a^2 + b^2 \equiv 2009 \equiv 4 (mod \: 5)$ . Now since $a^2 \equiv 0,1 \: or \: 4 (mod \: 5)\: and \: b^2 \equiv 0,1 \: or \: 4 (mod \: 5)$ , it implies that $a \equiv 0 \: or \: 4$ . Now also looking at the equation modulo 3, we get $a^2 + b^2 \equiv 2 \: (mod \: 3)$ . Therefore, $a^2 \equiv 1 (mod \: 3)$ and $a \equiv 1 \: or \: 2 (mod \: 3)$ . Hence in the interval (31 : 45), the possible values for a are 34,35,40,44. Of these four values the only value which satisfies the given equation is a = 35. $\square$

Start by looking at the units place. For it to be 9, the units place of a^2 and b^2 have to end in either (0,9) or (4,5). For this 'a' and 'b' have to end in either { 0, (3,7) } or { (2,8), 5 }. So if you just sort out the numbers below 45 ( 45^2=2025 ), you find that only (28,35) is the possible solution. As a>b, a=35.

Code

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for a in range(1,2009):
    for b in range(0,2009):
        if (a**2+b**2==2009):
            if (a>b):
                print(a,b)

Output

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35 28