system of equations: x + y = 7 y + x = 11Its pretty visible that the solution is ( x , y ) = ( 9 , 4 )For this, I put x = p 2 and y = q 2 . Then I subtracted one equation from the another such that I got 4 on RHS and factorized LHS to get two factors in terms of p and q.Then 4 can be represented as 2 ∗ 2, 4 ∗ 1 or 1 ∗ 4. Comparing the two factors on both sides, I got the solution.As you can see, the major drawback here is that I assumed this system has only integral solutions and then went further. Is there any way I can prove that this system indeed has only integral solutions or is there any other elegant way to solve this question?

Question

system of equations:      x    +  y  =  7      y    +  x  =  11Its pretty visible that the solution is   (  x  ,  y  )  =  (  9  ,  4  )For this, I put   x  =        p      2   and   y  =        q      2  . Then I subtracted one equation from the another such that I got   4 on RHS and factorized LHS to get two factors in terms of   p and   q.Then   4 can be represented as   2  ∗  2,   4  ∗  1 or   1  ∗  4. Comparing the two factors on both sides, I got the solution.As you can see, the major drawback here is that I assumed this system has only integral solutions and then went further. Is there any way I can prove that this system indeed has only integral solutions or is there any other elegant way to solve this question?

tatovihd1 · Accepted Answer

You want      {                            x          =          (          7          −          y                      )            2                                                y          =          (          11          −          x                      )            2                                                0          ≤          x          ≤          11                                      0          ≤          y          ≤          7                        The equation becomes  x  =  49  −  14  (  121  −  22  x  +      x    2    )  +  (  121  −  22  x  +      x    2        )    2  which reduces to  (  x  −  9  )  (      x    3    −  35      x    2    +  397  x  −  1444  )  =  0The polynomial   f  (  x  )  =      x    3    −  35      x    2    +  397  x  −  1444 has at least one real root. It has indeed three. One of them satisfies the condition   0  ≤  x  ≤  11 and it&#039;s approximately   7.87. With this value of   x we get   y  ≈  9.79 that doesn&#039;t satisfy   0  ≤  y  ≤  7.One can be more precise: call   α the least root of   f. Then   7  &amp;lt;  α  &amp;lt;  8, so that   3  &amp;lt;  11  −  α  &amp;lt;  4 and so   9  &amp;lt;  (  11  −  α      )    2    &amp;lt;  16, which shows that the limitation on   y is not fulfilled.How can you know this? Compute-   f  (  7  )  =  −  37-   f  (  8  )  =  4-   f  (  12  )  =  8-   f  (  13  )  =  −  1-   f  (  15  )  =  11Thus you know that the three roots of   f are   7  &amp;lt;  α  &amp;lt;  8,   12  &amp;lt;  β  &amp;lt;  13 and   13  &amp;lt;  γ  &amp;lt;  15.

system of equations: <msqrt> x </msqrt> + y = 7 <msqrt> y </

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