Squarimes (Square Primes)

Whilst getting out of bed, I was thinking of this:

(px)2+(px+n)2=(py)2(p_x)^2 + (p_{x + n})^2 = (p_y)^2 - where all variables are prime - (py)(p_y) is another prime.

The equation cannot be simplified algebraically.

We'll have to prove this numerically (i.e. manually).

Let's do the primes 20\leq 20:

22+32=132^2 + 3^2 = 13

22+52=292^2 + 5^2 = 29

32+52=343^2 + 5^2 = 34

22+72=532^2 + 7^2 = 53

32+72=563^2 + 7^2 = 56

52+72=745^2 + 7^2 = 74

22+92=852^2 + 9^2 = 85

32+92=903^2 + 9^2 = 90

52+92=1065^2 + 9^2 = 106

72+92=1307^2 + 9^2 = 130

22+112=1252^2 + 11^2 = 125

32+112=1303^2 + 11^2 = 130

52+112=1465^2 + 11^2 = 146

72+112=1707^2 + 11^2 = 170

22+132=1732^2 + 13^2 = 173

32+132=1783^2 + 13^2 = 178

52+132=1945^2 + 13^2 = 194

72+132=2187^2 + 13^2 = 218

112+132=29011^2 + 13^2 = 290

22+172=2932^2 + 17^2 = 293

32+172=2983^2 + 17^2 = 298

52+172=3145^2 + 17^2 = 314

72+172=3387^2 + 17^2 = 338

112+172=41011^2 + 17^2 = 410

132+172=45813^2 + 17^2 = 458

22+192=3652^2 + 19^2 = 365

32+192=3703^2 + 19^2 = 370

52+192=3865^2 + 19^2 = 386

72+192=4107^2 + 19^2 = 410

112+192=48211^2 + 19^2 = 482

132+192=53013^2 + 19^2 = 530

172+192=65017^2 + 19^2 = 650

Now looking at the square numbers (up to limit of what we've done):

1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484,529,576,6251, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625

Now reducing it to prime square numbers (i.e. when primes are squared, it produces a square number):

4,9,25,49,81,121,169,289,361,5294, 9, 25, 49, 81, 121, 169, 289, 361, 529

The closest squarime is 132+192=53013^2 + 19^2 = 530 which is 11 away from the nearest square number, that is 529529.

I don't know if there is any squarimes defined like this.

I'd request anybody to find a general function that proves whether there is or not any squarimes according to my definitions and restrictions.

Note by A Former Brilliant Member
1 month, 4 weeks ago

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@Yajat Shamji- There never exists a2+b2=c2a^2+b^2=c^2 such that all of them are odd. \because any set of integers satisfying the equation (a2+b2=c2a^2+b^2=c^2) can be represented in the form a=2mn;b=m2n2;c=m2+n2;mZ;nZa=2mn;b=m^2-n^2;c=m^2+n^2;m \in Z; n \in Z as you can see aa is even \therefore all of them can't be prime.

Also if you see if a,ba,b are odd then cc must be even \therefore all of them can never be prime

Zakir Husain - 1 month, 4 weeks ago

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Thank you so much! @Zakir Husain

A Former Brilliant Member - 1 month, 4 weeks ago

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What @Zakir Husain said can be backed by brute force as well

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import math

checkMe = range(1, 10000)
primes = []
for y in checkMe[1:]:
    x = y
    dividers = []
    for x in range(2, x):
        if (y/x).is_integer():
            dividers.append(x)
    if len(dividers) < 1:
        primes.append(y)

for i in primes:
    for j in primes:
        print(math.sqrt(i**2 + j**2))

1

Mahdi Raza - 1 month, 4 weeks ago

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Thank you so much! @Mahdi Raza

A Former Brilliant Member - 1 month, 4 weeks ago

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@Yajat Shamji- See a more elaborated proof here

Zakir Husain - 1 month, 4 weeks ago

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Thank you so much! @Zakir Husain

A Former Brilliant Member - 1 month, 4 weeks ago

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