Question

Show that the sum of squares of two odd integers cannot be the square of an integer.

Short answer

a is not a perfect square

Step-by-step solution

Step 1

$a=(2m+1{)}^2+(2n+1{)}^2=4\left(m^2+n^2+m+n\right)+2\equiv \mathrm{mod}4$

a is an even integer. If a were to be a perfect square, then it should be divisible by 4 (since it divisible by 2 ).

But that is not the case as seen from above.

Hence is not a perfect square.