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If I understand correctly, then A=1130 is a counter example. A^2+1 = 1276901 = 577*2213, composite. The only decompositions of A^2+1 into a sum of two squares are 95^2+1126^2 and 1^2+1130^2 so the best b is b=95 and 10*b=950<1130=A.

Unless I am off track, each of the following values of A is also a counterexample:
1130, 1624, 2284, 2630, 3296, 3486, 4036, 4640, 5050, 5746, 6376, 6706, 6744, 7124, 7746, 8036, 8040, 8950, 9604

A more extreme counterexample is A=83030, where largest b=815, so 100*b<A....Show more

a^2+b^2 = A^2+1

so doesn't b = (+/-) 1

Am I missing something to your question?

Then let a = 0, a^2+b^2 = 1 not composite
a = 2, a^2 + b^2 = 5, not composite
a = 4, a^2 + b^2 = 17 not composite
a = 6, a^2 + b^2 = 37, not composite
a = 8, a^2 + b^2 = 65 = 13*5 composite

a = 8

10*1 > 8

So If I am reading this problem correctly, here is a solution.

But the question was can you ALWAYS.

I have found counterexamples, so the universal is disproven.

But I have also shown existance....Show more