Mysterious Square

Let's look at these links:

//leitura dos elementos
for(i = 0; i < n; i++){
    for(j = 0; j < n; j++){
        scanf("%d", &elem);
        //somatorio de cada linha
        aux += i;
        soma_linhas = aux;
    }
}

Regardless of what happens, at the end soma_linhas and aux will have the same value. What will be the value of aux . For a square of 3x3, it will be 0 in the first iteration, 1 in the second (+1) and 3 in the third (+2). However, that was not what he should do! There is no point in adding the values of i and disregarding the values of elem . You should check the values of elem .

However, even if you use aux += elem; , it will still be wrong, as this will end up adding all the values, not the values of each line separately.

What you have to do is to change this loop so that it adds up the values of each line and from the second line, compare with the value of the previous line.

Note that you are not storing the read values anywhere, the variable elem will only save the last value. Because of this, your loop to check the main diagonal will not work as you are only adding fixed values 0, 1 and 2 and completely ignoring the contents of the array.

Without storing the entire array, it is very difficult to verify that all rows and the main diagonal have the same values. But since you can not store values in vectors, you do the following:

• Use a variable (let's call ok ) to indicate whether the sums of the lines and the main diagonal match. Initialize it with 1.

• Read the first line, adding up all the items (let's call it a ). Store the first element in an auxiliary variable ( b ).

• Use for to read the other lines, adding the elements of each line to another variable c . The main column element also adds the b variable. At the end of each line, check if a == c and change ok to 0 if it is not. Do not forget to return c to 0 before starting the next line.

• When you finish the last line, check if a == b and change ok to 0 if it is not.

• At the end, just check the variable ok .

Something like this:

#include <stdio.h>

int main() {

    int n;
    scanf("%d", &n);

    int elem, a = 0, b = 0, ok = 1;
    for (int i = 0; i < n; i++) {
        int c = 0;
        for (int j = 0; j < n; j++) {
            scanf("%d", &elem);
            if (i == j) b += elem;
            if (i == 0) a += elem;
            c += elem;
        }
        if (a != c) ok = 0;
    }
    if (a != b) ok = 0;

    if (ok) {
        printf("Quadrado misterioso com soma %d.", a);
    } else {
        printf("Não é um quadrado misterioso.");
    }
    return 0;
}

See working in ideone:

• For a mysterious square.

You do not mention in the question anything about the sum of the columns, according to Wikipedia the definition of a Quadrado Mágico is:

  

Magic Square is a square table equal to the intersection of numbers   in which the sum of each column, each row and the two diagonals are   equal.

     

Here'satestedandcommentedsolutiontoseeifamysterioussquareismagicalornot,see:

#include<stdlib.h>#include<stdio.h>inteh_magico(int**quad,intn){inti=0,j=0;intx=0,y=0;intd1=0,d2=0;/*Somatoriodasdiagonais*/for(i=0;i<n;i++){d1+=quad[i][i];d2+=quad[n-i-1][i];}/*Verificasediagonaissaoiguais*/if(d1!=d2)return-1;for(i=0;i<n;i++){x=0;y=0;/*Somatoriolinhaecoluna*/for(j=0;j<n;j++){x+=quad[i][j];y+=quad[j][i];}/*Verificalinhaecoluna*/if(x!=d1||y!=d1)return-1;}/*Ehmagico!*/returnd1;}intmain(void){intlin,col,dim;int**quadrado_misterioso;/*Ledimensoes*/scanf("%d", &dim );

    /* Aloca memoria para n linhas */
    quadrado_misterioso = malloc( dim * sizeof(int*) );

    /* Para cada linha */
    for( lin = 0; lin < dim; lin++ )
    {
        /* Aloca memoria para n colunas da linha atual */
        quadrado_misterioso[lin] = malloc( dim * sizeof(int) );

        /* Para cada coluna da linha */
        for( col = 0; col < dim; col++ )
            scanf( "%d", &quadrado_misterioso[lin][col] );
    }

    /* Verifica se eh um quadrado magico */
    printf("%d\n", eh_magico( quadrado_misterioso, dim ) );

    /* Libera memoria ocupada pelo quadrado */
    for( lin = 0; lin < dim; lin++ )
        free(quadrado_misterioso[lin]);
    free(quadrado_misterioso);

    return 0;
}

Test (% with%):

3
4   9   2
3   5   7
8   1   6
15

Test (% with%):

4
2   16  13  3
11  5   8   10
7   9   12  6
14  4   1   15
34

Test (% with%):

5
1   23  16  4   21
15  14  7   18  11
24  17  13  9   2
20  8   19  12  6
5   3   10  22  25
65

Test (% with%):

3
1   2   3
4   5   6
7   8   9
-1

Test (% with%):

3
2   7   11
9   5   6
4   3   13
-1

See working at Ideone.com