Problem 4

The positive integers

a

and

b

are such that

15a+16b

and

16a-15b

are both squares of positive integers. What is the least possible value of the smaller of these two squares?

Step 4 of 4: Attain the bound

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In plain words

Attain the bound

\min(m^2,n^2)\ge481^2

Detailed analysis

Write $m=481m'$ and $n=481n'$ . Since $m',n'\ge1$ , the smaller square is at least $481^2$ . Taking $m'=n'=1$ gives $a=481(15+16)=481\cdot31$ and $b=481(16-15)=481$ , for which both original expressions equal $481^2$ .