Divide 29 into two parts so that the sum of the squares of the parts is 425.

Solution:

Let first numbers be $x$ and other $(29-x)$

Then according to question

$x^{2}+(29-x)^{2}=425$

$x^{2}+x^{2}-58 x+841=425$

$2 x^{2}-58 x+841=425$

$2 x^{2}-58 x+841-425=0$

$2 x^{2}-58 x+416=0$

$x^{2}-29 x+208=0$

$x^{2}-16 x-13 x+208=0$

$x(x-16)-13(x-16)=0$

$(x-16)(x+13)=0$

$(x-16)=0$

$x=16$

Or

$(x+13)=0$

$x=-13$

Since, 29 being a positive number, so x cannot be negative.

Therefore,

When $x=16$ then

$29-x=29-16$

$=13$

Thus, two consecutive number be