Check that number is equal to the sum of the squares of 4 consecutive prime numbers

If you do a search on the website you will find several different algorithms to determine if a number is Prime (If you search on Google then you will find many others, and you will see that your algorithm is far from correct).

But even if it was correct, in the end you sum num ** 2 with (num + 1) ** 2, and this is already wrong. If num is prime, then num + 1 is Prime only if num equals 2. For any other prime number, num + 1 will be an even number (and therefore, divisible by 2, and therefore will not be prime). The same goes for num + 2 and num + 3, at least one of them will not be prime.

There is actually no fixed interval between prime numbers. Given any prime number, the next prime number can be any distance ahead (on this assumption I suggest a read here), so there is no way to add 1 (or any other fixed value) and surely find another prime number. The way is to calculate it even.

^{in this comment you said You can't use lists or functions ( which is a restriction that should be in the question , so people don't waste time writing answers that don't serve can focus on the most suitable solution for your case).}

Without using lists or functions, a solution would be to save 4 consecutive prime numbers in variables and check if the sum of their squares is equal to number in question. If not, I calculate the next prime number, do the sum again, and so it goes.

import math
n = int(input("n: "))

# já inicio com os 4 primeiros primos
p1, p2, p3, p4 = 2, 3, 5, 7
while True:
    if (p1 ** 2) + (p2 ** 2) + (p3 ** 2) + (p4 ** 2) == n:
        print(f'Encontrei: {p1}, {p2}, {p3} e {p4}')
        break # encontrei, sai do loop

    # não encontrei, buscar o próximo primo
    proximo = p4 + 2 # posso começar a partir de p4
    while proximo <= n: # loop para verificar se "proximo" é primo
        # loop pode ir até raiz quadrada do número: https://stackoverflow.com/q/5811151 
        # range de 2 em 2, para só testar números ímpares
        # como "proximo" é ímpar, não preciso começar o range em 2, pode começar direto no 3
        for i in range(3, int(math.sqrt(proximo) + 1), 2):
            if proximo % i == 0:
                primo = False
                break # "proximo" não é primo, sai do for
        else: # se o for não for interrompido pelo break, cai nesse else
            primo = True

        if primo: # atualiza a lista de primos e sai do while interno (volta para o while externo, que verifica a soma dos quadrados)
            p1, p2, p3, p4 = p2, p3, p4, proximo
            break
        else: # não é primo, tentar o próximo número ímpar
            proximo += 2

    if proximo > n: # se já passou de n, é porque não tem
        print(f'{n} não é igual a soma de 4 primos consecutivos ao quadrado')
        break # sai do loop

If you could use lists, an alternative would be to have a list with all the primes up to n, and then run through it doing the check (as did another answer, for example).

The algorithm below was taken from this response , and is an implementation of the Eratosthenes sieve :

import math
n = int(input("n: "))

# criar lista com todos os primos até n (algoritmo do Crivo de Eratóstenes)
primos = list(range(2, n))
for i in range(2, int(math.sqrt(n) + 1)):
  if i in primos:
    for j in range(i ** 2, n, i):
      if j in primos:
        primos.remove(j)

# verificar a soma dos quadrados
for i in range(len(primos) - 3):
    if (primos[i] ** 2) + (primos[i + 1] ** 2) + (primos[i + 2] ** 2) + (primos[i + 3] ** 2) == n:
        print(f'Encontrei: {primos[i]}, {primos[i + 1]}, {primos[i + 2]} e {primos[i + 3]}')
        break # encontrei, sai do loop
else:
    print(f'{n} não é igual a soma de 4 primos consecutivos ao quadrado')

Evident that gives to optimize, as you don't need to create the list with all the primes up to n, it could go to the square root of n divided by 2 , among other details. But for one exercise, it should already be enough.

Tries:

def ehprimo(num,primos):
    for divi in primos:
        if num%divi == 0:
            return False
    return True

num,cont,primos = 2,1,[]
while cont <= 100000:
    if ehprimo(num,primos) == True:
        cont += 1
        print(num)
        primos.append(num)
    num += 1
n = int(input("Informe um número: "))
i = 0
while i < len(primos)-3:
    if n == (primos[i]**2 + primos[i+1]**2 + primos[i+2]**2 + primos[i+3]**2):
        print(n, " = ", primos[i], "^2 + ", primos[i+1], "^2 + ", primos[i+2], "^2 + ", primos[i+3], "^2")
        break
    i += 1