Find the sum of positive integers x and y, satisfying x^2+y^2=8768.

The video illustrates step-by-step how to find the sum of positive integers x and y, satisfying x^2+y^2=8768. In general, to find these positive integers, it’s useful to narrow down the range of possible integers (e.g. by checking the parity of the variables), especially for larger numbers.

However, an effective method for solving this kind of problems may vary from problem to problem. For this problem, noticing 8768 is divisible by 4 is vital to solving the problem.

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[owlsmath]

Nice problem! I used the prime factorization to write it as 64*137. 64 = 8^2+0^2, 137 = 11^2 + 4^2. Then 8*11=88 and 8*4=32 and 88+32=120

[sy]

For the additional problem: My answer is 76, 92, 100

{At first, "both u and v" are "even numbers" or "odd numbers".}
(2k+1)^2=4(k^2+k)+1=4k'+1 (k, k': integer) Therefore (odd number)^2+(odd number)^2[=4k''+2] cannot be divided by 4.
5200/4=1300 This means 4 | u^2+v^2 . So, "both u and v" are "even numbers". Set u=2m, v=2n, (2m)^2+(2n)^2=5200 . ∴ m^2+n^2=1300 ( 1300/4=325 )

Next to this, from the same theory, after setting m=2s, 2t, we get (2s)^2+(2t)^2=1300 ∴ s^2+t^2=325
If s≧t, s^2≧t^2(>0), and from this s^2≧325/2=162.5 . By the way, "13^2=169, 12^2=144", and"19^2=361, 18^2=324". ∴ 13≦s≦18 (∵ s, t: integers)
If s=18, t^2=325-18^2=325-324=1 ∴ t=1 (∵ t>0)
If s=17, t^2=325-17^2=325-289=36 ∴ 6=1 (∵ t>0)
If s=16, t^2=325-16^2=325-256=69 t is NOT perfect square.
If s=15, t^2=325-15^2=325-225=100 ∴ t=10 (∵ t>0)
If s=14, t^2=325-14^2=325-196=129 t is NOT perfect square.
If s=13, t^2=325-13^2=325-169=156 t is NOT perfect square.

∴ (s, t) = (18, 1), (17, 6), (15, 10) ∴ (m, n) = (36, 2), (34, 12), (30, 20) ∴ (u, v) = (72, 4), (68, 24), (60, 40) [ u and v are symmetrical: (4, 72), (24, 68), (40, 60) ]
∴ u+v = (72+4=)76, (68+24=)92, (60+40=)100

Is my method (and calculation) right?
[Yesterday's additional problem was difficult, but today's one is not difficult. So, I'm relieved. ]

[vcvartak]

60, 40, 100 are answers.
60**2+40**2=5200
You can reduce problem to x**2 + y**2 = 52 and then multiply x, y by 10.

[ABHISHEKKUMAR-]

Ma'm, with your permission I am proceeding thus :
Supposing u = 2p and v = 2q, we get
p^2 + q^2 = 1300
Again, supposing p = 2r and q = 2s, we get
r^2 + s^2 = 325
Now, considering r > s > 0, we get
r^2 < 325 < 2r^2
Now, r^2 < 325 => 1 < r < 18, where the numbers 1 and 18 are included
And, 2r^2 > 325
i.e. r^2 > 162.5
which => r > 13, where the number 13 is included
Combining the two conditions, we get
13 < r < 18 (Numbers 13 and 18 are included)
By substituting r = 13, 14, ..., 18, we find that there are three pairs (r, s) = (15, 10), (17, 6) and (18, 1) which satisfies the relation r^2 + s^2 = 325
Now, (r, s) = (15, 10), (17, 6) and (18, 1)
=> (p, q) = (30, 20), (34, 12) and (36, 2)
And, (u, v) = (60, 40), (68, 24) and (72, 4)
Hence three possible values of u + v are
u + v = 60 + 40, 68 + 24 and 72 + 4
i.e. u + v = 100, 92 and 76.

[chrismcgowan]

problem 1, x + y = 120, problem 2, u + v = 76, 92, 100