# Sum of squares of divisors

Let $$n$$ be a natural number.

Suppose $$d_{1},d_{2},d_{3}, \ldots$$ are the divisors of $$n$$ such that

$$1=d_{1}<d_{2}<d_{3}<\ldots$$

Suppose $$n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$$

Find the value of $$n$$.

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