This challenge was very simple. If you analyze the hashing code, it only uses the lower image dimension for computing the hash.
So the simplest solution is uploading two 1x2 pictures:

(black, white - black, black)

```
import math
import hashlib
from PIL import Image

def HashImage(image):
def md5(fname):
hash_md5 = hashlib.md5()
with open(fname, "rb") as f:
for chunk in iter(lambda: f.read(4096), b""):
hash_md5.update(chunk)
return hash_md5.hexdigest()

im=image
pix=im.load()
width, height = im.size
small=height if width>height else width
pic=[]

# picture 1x2:
# small = 1 !!!!!!!!!!!!!!!!!!!!!!!!!!!!

for i in range(small):
curr=[]
for j in range(i,small):
curr.append(pix[i,j])
for j in range(i+1,small):
curr.append(pix[j,i])
sum,mul,sub=0,1,1000000

if i%3==0:
for pixel in curr:
sum+=pixel[0]
mul*=pixel[1]
sub-=pixel[2]

elif i%3==1:
for pixel in curr:
sum+=pixel[2]
mul*=pixel[0]
sub-=pixel[1]
else:
for pixel in curr:
sum+=pixel[1]
mul*=pixel[2]
sub-=pixel[0]

pic.append((sum%256,mul%256,sub%256))

size=int(math.floor(math.sqrt(len(pic))))
pic=pic[0:size*size]

im2 = Image.new('RGB', (size,size),'white')
im2.putdata(pic)
im2.save("hash.jpg")
return md5('hash.jpg')
```

The flag:
```
evlz{md5_is_lub}ctf
```