The Diophantine equation:

**(x**

^{2}+ ay^{2})(u^{2}+ bv^{2}) = p^{2}+ (a + b +**ab**

**)q**

^{2}will always have an infinity of nonzero integer solutions which you can find using BQD Iterators.

Thought it might be fun to consider some recent sums of distinct squares I've given to play with the result.

Let a = 4 + 9 = 13, and b = 4 + 9 + 25 + 49 = 87, so I can find where I'm pulling sums of squares from prior posts:

(

**92**)(

^{2}+ 96^{2}+ 144^{2}**7048**) =

^{2}+ 688^{2}+ 1032^{2 }+ 1720^{2 }+ 2408^{2}**p**

^{2}+ 1231**q**

^{2}

Matched by iterations which is necessary and used the third iteration, so p and q should be easy to calculate.

The BQD iterator is:

Given nonzero integers u and v with

u

^{2}+ 1231v

^{2}= F

then it must also be true that

(u - 1231v)

^{2}+ 1231(u + v)

^{2}= 1232*F

Where u = v = 1 to start will give F = 1232.

1

^{2}+ 1231*1

^{2}= 1232

First iteration: (-1230)

^{2}+ 1231*(2)

^{2}= 1232

^{2}

Second iteration: (-3692)

^{2}+ 1231*(-1228)

^{2}= 1232

^{3}

Third iteration: (1507976)

^{2}+ 1231*(-4920)

^{2}= 1232

^{4}

Which means that p = 1507976, and q = 4920 are solutions, taking the positive values.

**(92**

^{2}+ 96^{2}+ 144^{2})(7048^{2}+ 688^{2}+ 1032^{2 }+ 1720^{2 }+ 2408^{2}) = 1507976^{2}+ 1231*4920^{2}Of course now I wonder what is the smallest number of distinct squares necessary to sum to 1231. Please feel free to reply with an answer in the comments. Doubt I'll work at figuring that one out.

Got bored, found 26

^{2}+ 23

^{2}+ 5

^{2}+ 1

^{2}= 1231

Guess I should show that then:

**(92**

^{2}+ 96^{2}+ 144^{2})(7048^{2}+ 688^{2}+ 1032^{2 }+ 1720^{2 }+ 2408^{2}) =

**1507976**

^{2}+ 127920^{2}+ 113160^{2}+ 24600^{2}+ 4920^{2}Doesn't mean much to me. And don't know if that's the smallest number of distinct squares that will work.

My mind will just start playing with numbers whether I like it or not, almost will just go on automatic. That doesn't sound normal to me. But I'm ways away from anything close to normal I guess.

James Harris

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