Find a number of the form which is divisible by **5**,

the quotient being a square.

If we take x = 5, and y equal to a square such that and are also squares.

The least possible value for y is 4, in which case

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Let’s look at ordered triples of integers whose squares form an arithmetic progression.

In other words,

, or equivalently

The solutions are of the form

for any integers and

The ordered triple is a primitive arithmetic progression triple if

We would need to have:

that is,

if m = 5 and n = 4, then

a = -31, b = 41, c = 49

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See also Paul’s solutions at the bottom of the page

Here are a few, format is {x, y, 4 x y(x + y)(x – y), Sqrt[(4 x y(x + y)(x – y))/5]}

{5,4,720,12}

{9,1,2880,24}

{10,8,11520,48}

{15,12,58320,108}

{18,2,46080,96}

{20,16,184320,192}

{25,20,450000,300}

{27,3,233280,216}

{30,24,933120,432}

{35,28,1728720,588}

{36,4,737280,384}

{40,32,2949120,768}

{45,5,1800000,600}

{45,36,4723920,972}

{50,40,7200000,1200}

{54,6,3732480,864}

{55,44,10541520,1452}

{60,48,14929920,1728}

{63,7,6914880,1176}

{65,52,20563920,2028}

{70,56,27659520,2352}

{72,8,11796480,1536}

{75,60,36450000,2700}

{80,64,47185920,3072}

{81,9,18895680,1944}

{85,68,60135120,3468}

{90,10,28800000,2400}

{90,72,75582720,3888}

{95,76,93831120,4332}

{99,11,42166080,2904}

{100,80,115200000,4800}

Paul.

Thanks. I posted a different solution