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Discounting one (Posted on 2018-01-10) Difficulty: 3 of 5
Any three of the integers {1, X, Y, Z} add up to a perfect square.

What are X, Y, & Z?

Hint: each is below 100.

No Solution Yet Submitted by Ady TZIDON    
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An interesting pattern | Comment 8 of 9 |
Let 1<X<Y<Z 

We can call m, n, p, q the integers whose squares are the sums of each one of the three groups (i.e: 1+X+Y=m^2, 1+X+Z=n^2, and so on). For easyness I will call M, N, P, Q the squares of m, n, p, q, respectively

Is obvius that: 3+2Q=M+N+P

As the first half of the equality is odd, the solutions should be only with m,n,p (or M, N, P) all three odds or with one odd and the other two even numbers.

It is also very easy to obtain the values of X,Y,Z in terms of m,n,p (or M, N, P)

X= (M+N-P-1)/2
Y= (M-N+P-1)/2
Z= (-M+N+P-1))/2
Using this there is an interesting pattern that I discover accidentaly. See the table below: 

 q        m   n    p         X       Y       Z
 0        1    1    1         0       0       0
 6        5    5    5        12    12      12
11       8    9   10        22    41      58
16     11   13   15       32    88     136
21     14   17   20       42   153    246
26     17   21   25

As we can observ for each ten values of X there is a solution (at least)
That solution is related to each five values of Q, and X=2Q
Between M- N and N-P there is always the same diference, and the difference increases one unity at every solution. 

I don't know the reason of this but it is a beautiful pattern!

Edited on January 12, 2018, 2:36 pm
  Posted by armando on 2018-01-12 11:23:51

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