Squares of squares

Let $4m$ and $3m$ be the side lengths of the initial two squares and $n$ be the side length of the square of intersection. The total area showing will then be the sum of the areas of the initial two squares minus that of the square of intersection. Thus we have that

$(4m)^{2} + (3m)^{2} - n^{2} = 5000 \Longrightarrow 25m^{2} - n^{2} = 5000 \Longrightarrow n^{2} = 25(m^{2} - 200)$ .

Now as $25|n^{2}$ we must have that $5|n$ , so let $n = 5k$ . This last equation then becomes

$(5k)^{2} = 25(m^{2} - 200) \Longrightarrow k^{2} = m^{2} - 200 \Longrightarrow m^{2} - k^{2} = 200 \Longrightarrow (m - k)(m + k) = 200$ .

Now if $m - k = a$ and $m + k = b$ then $m = \dfrac{a + b}{2}$ and $k = \dfrac{b - a}{2}$ . So in order for $m,k$ to be integers we will require that either both of $a,b$ are even or both odd. Given that $200 = 2^{3} \times 5^{2}$ , the possible values for $(a,b)$ are $(2,100), (4,50)$ and $(10,20)$ , yielding respective values for $(m,k)$ of $(51,49), (27,23)$ and $(15,5)$ . But as we require that $n = 5k \lt 4m$ the only option for $(m,k)$ that is suitable is $(15,5)$ , which means that the original two squares have side lengths $60$ and $45$ , while the square of intersection has side length $25$ . The side length of the largest square is thus $\boxed{60}$ .