# Squarimes (Square Primes)

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

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

The equation cannot be simplified algebraically.

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

Let's do the primes $\leq 20$:

$2^2 + 3^2 = 13$

$2^2 + 5^2 = 29$

$3^2 + 5^2 = 34$

$2^2 + 7^2 = 53$

$3^2 + 7^2 = 56$

$5^2 + 7^2 = 74$

$2^2 + 9^2 = 85$

$3^2 + 9^2 = 90$

$5^2 + 9^2 = 106$

$7^2 + 9^2 = 130$

$2^2 + 11^2 = 125$

$3^2 + 11^2 = 130$

$5^2 + 11^2 = 146$

$7^2 + 11^2 = 170$

$2^2 + 13^2 = 173$

$3^2 + 13^2 = 178$

$5^2 + 13^2 = 194$

$7^2 + 13^2 = 218$

$11^2 + 13^2 = 290$

$2^2 + 17^2 = 293$

$3^2 + 17^2 = 298$

$5^2 + 17^2 = 314$

$7^2 + 17^2 = 338$

$11^2 + 17^2 = 410$

$13^2 + 17^2 = 458$

$2^2 + 19^2 = 365$

$3^2 + 19^2 = 370$

$5^2 + 19^2 = 386$

$7^2 + 19^2 = 410$

$11^2 + 19^2 = 482$

$13^2 + 19^2 = 530$

$17^2 + 19^2 = 650$

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

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

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

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

The closest squarime is $13^2 + 19^2 = 530$ which is $1$ away from the nearest square number, that is $529$.

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 year, 1 month ago

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- 1 year, 1 month ago

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

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

- 1 year, 1 month ago

Thank you so much! @Zakir Husain

- 1 year, 1 month ago

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

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 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  

- 1 year, 1 month ago

Thank you so much! @Mahdi Raza

- 1 year, 1 month ago

@Yajat Shamji- See a more elaborated proof here

- 1 year, 1 month ago

Thank you so much! @Zakir Husain

- 1 year, 1 month ago